{"id":33935,"date":"2022-01-25T17:55:51","date_gmt":"2022-01-25T12:25:51","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=33935"},"modified":"2024-03-08T11:54:21","modified_gmt":"2024-03-08T06:24:21","slug":"important-questions-for-class-10-maths-chapter-10-circles","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-10\/circles\/","title":{"rendered":"Important Questions for Class 10 Maths Chapter 10 Circles"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-10\/circles\/#Class_10_Important_Questions_Circles_-_Very_Short_Answer_1_Mark\" title=\"Class 10 Important Questions Circles &#8211; Very Short Answer (1 Mark)\">Class 10 Important Questions Circles &#8211; Very Short Answer (1 Mark)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-10\/circles\/#Circles_Class_10_Important_Questions_Short_Answer-I_2_Marks\" title=\"Circles Class 10 Important Questions Short Answer-I (2 Marks)\">Circles Class 10 Important Questions Short Answer-I (2 Marks)<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-10\/circles\/#Circles_Class_10_Important_Questions_Short_Answer-II_3_Marks\" title=\"Circles Class 10 Important Questions Short Answer-II (3 Marks)\">Circles Class 10 Important Questions Short Answer-II (3 Marks)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-10\/circles\/#Circles_Class_10_Important_Questions_Long_Answer_4_Marks\" title=\"Circles Class 10 Important Questions Long Answer (4 Marks)\">Circles Class 10 Important Questions Long Answer (4 Marks)<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<p>In Chapter 10 of Class 10 Maths, titled &#8220;Circles,&#8221; several important concepts are explored. One crucial aspect is understanding the properties and theorems related to circles. Key questions might revolve around the relationships between different parts of a circle, such as the radius, diameter, chord, and arc. Theorems like the Inscribed Angle Theorem and the Angle Subtended by an Arc at the Center are fundamental for solving problems related to circles. Additionally, questions on tangents, secants, and their properties within a circle are essential to grasp. To excel in this chapter, students should focus on mastering the theorems, practicing geometric constructions involving circles, and solving numerical problems that test their comprehension of circle-related concepts.<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/\" target=\"_blank\" rel=\"noopener\">CBSE Class 10 Maths Important Question Chapter 8 Tribometry<\/a><\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Class_10_Important_Questions_Circles_-_Very_Short_Answer_1_Mark\"><\/span>Class 10 Important Questions Circles &#8211; Very Short Answer (1 Mark)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1. In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If \u2220AOB = 100\u00b0, then calculate \u2220BAT. (2011D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130413\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-1.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 1\" width=\"107\" height=\"101\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130414\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-2.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 2\" width=\"121\" height=\"113\" \/><br \/>\n\u22201 = \u22202<br \/>\n\u22201 + \u22202 + 100\u00b0 = 180\u00b0<br \/>\n\u22201 + \u22201 = 80\u00b0<br \/>\n\u21d2 2\u22201 = 80\u00b0<br \/>\n\u21d2 \u22201 = 40\u00b0<br \/>\n\u22201 + \u2220BAT = 90\u00b0<br \/>\n\u2220BAT = 90\u00b0 \u2013 40\u00b0 = 50\u00b0<\/p>\n<p style=\"text-align: center;\"><strong>Also Check:<a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/important-questions-for-class-10-maths-chapter-14-statistics\/\" target=\"_blank\" rel=\"noopener\"> Important Questions for Class 10 Maths Chapter 14 Statistics<\/a><\/strong><\/p>\n<p><strong>Question 2. In the given figure, PA and PB are tangents to the circle with centre O. If \u2220APB = 60\u00b0, then calculate \u2220OAB, (2011D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130416\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-3.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 3\" width=\"148\" height=\"108\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130417\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-4.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 4\" width=\"156\" height=\"120\" \/><br \/>\n\u22201 = \u22202<br \/>\n\u22201 + \u22202 + \u2220APB = 180\u00b0<br \/>\n\u22201 + \u22201 + 60\u00b0 = 180\u00b0<br \/>\n2\u22201 = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br \/>\n\u22201 = \\(\\frac{120^{\\circ}}{2}\\) = 60\u00b0<br \/>\n\u22201 + \u2220OAB = 90\u00b0<br \/>\n60\u00b0 +\u2220OAB = 90\u00b0<br \/>\n\u2220OAB = 90\u00b0 \u2013 60\u00b0 = 30\u00b0<\/p>\n<p><strong>Question 3. In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If \u2220POQ = 70\u00b0, then calculate \u2220TPQuestion (2011OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130418\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-5.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 5\" width=\"147\" height=\"145\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130419\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-6.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 6\" width=\"147\" height=\"132\" \/><br \/>\n\u22201 = \u22202<br \/>\n\u22201 + \u22202 + 70\u00b0 = 180\u00b0<br \/>\n\u22201 + \u22201 = 180\u00b0 \u2013 70\u00b0<br \/>\n2\u22201 = 110\u00b0 \u21d2 \u22201 = 55\u00b0<br \/>\n\u22201 + \u2220TPQ = 90\u00b0<br \/>\n55\u00b0 + \u2220TPQ = 90\u00b0<br \/>\n\u21d2 \u2220TPQ = 90\u00b0 \u2013 55\u00b0 = 35\u00b0<\/p>\n<p style=\"text-align: center;\"><strong>Also Check:<a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/surface-areas-and-volumes\/class-10\/extra-questions\/maths\/chapter-13\/\"> Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13<\/a><\/strong><\/p>\n<p><strong>Question 4. A chord of a circle of radius 10 cm subtends a right angle at its centre. Calculate the length of the chord (in cm). (2014OD)<\/strong><br \/>\nSolution:<br \/>\nAB<sup>2 <\/sup>= OA<sup>2<\/sup> + OB<sup>2<\/sup> \u2026[Pythagoras\u2019 theorem<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130420\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-7.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 7\" width=\"119\" height=\"119\" \/><br \/>\nAB<sup>2<\/sup> = 10<sup>2<\/sup> + 10<sup>2<\/sup><br \/>\nAB<sup>2<\/sup> = 2(10)<sup>2<\/sup><br \/>\nAB = \\(10 \\sqrt{2}\\) cm<\/p>\n<p>Question 5.<br \/>\nIn the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and \u2220CAB = 30\u00b0. Find \u2220PCA. (2016OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130421\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-8.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 8\" width=\"184\" height=\"124\" \/><br \/>\nSolution:<br \/>\n\u2220ACB = 90\u00b0 \u2026[Angle in the semi-circle<br \/>\nIn \u2206ABC,<br \/>\n\u2220CAB + \u2220ACB + \u2220CBA = 180\u00b0<br \/>\n30 + 90\u00b0 + \u2220CBA = 180\u00b0<br \/>\n\u2220CBA = 180\u00b0 \u2013 30\u00b0 \u2013 90\u00b0 = 60\u00b0<br \/>\n\u2220PCA = \u2220CBA \u2026[Angle in the alternate segment<br \/>\n\u2234 \u2220PCA = 60\u00b0<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/polynomials-extra-questions-maths-chapter-2\/\" target=\"_blank\" rel=\"noopener\">Polynomials Class 10 Extra Questions Maths Chapter 2<\/a><\/strong><\/p>\n<p><strong>Question 6. In the given figure, AB and AC are tangents to the circle with centre o such that \u2220BAC = 40\u00b0. Then calculate \u2220BOC. (2011OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130422\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-9.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 9\" width=\"166\" height=\"114\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130423\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-10.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 10\" width=\"166\" height=\"115\" \/><br \/>\nAB and AC are tangents<br \/>\n\u2234 \u2220ABO = \u2220ACO = 90\u00b0<br \/>\nIn ABOC,<br \/>\n\u2220ABO + \u2220ACO + \u2220BAC + \u2220BOC = 360\u00b0<br \/>\n90\u00b0 + 90\u00b0 + 40\u00b0 + \u2220BOC = 360\u00b0<br \/>\n\u2220BOC = 360 \u2013 220\u00b0 = 140\u00b0<\/p>\n<p><strong>Question 7. In the given figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K&amp;M respectively. If EK = 9 cm, calculate the perimeter of AEDF (in cm). (2012D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130424\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-11.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 11\" width=\"135\" height=\"175\" \/><br \/>\nSolution:<br \/>\nPerimeter of \u2206EDF<br \/>\n= 2(EK) = 2(9) = 18 cm<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/some-applications-of-trigonometry-class-10-extra-questions-maths-chapter-9\/\" target=\"_blank\" rel=\"noopener\">Some Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9<\/a><\/strong><\/p>\n<p><strong>Question 8. In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP (in cm). (2012OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130425\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-12.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 12\" width=\"134\" height=\"179\" \/><br \/>\nSolution:<br \/>\n2AP = Perimeter of \u2206<br \/>\n2AP = 5 + 6 + 4 = 15 cm<br \/>\nAP = \\(\\frac{15}{2}\\) = 7.5 cm<\/p>\n<p><strong>Question 9. In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA \u22a5 PB, then find the length of each tangent. (2013D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130426\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-13.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 13\" width=\"131\" height=\"122\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130427\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-14.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 14\" width=\"144\" height=\"153\" \/><br \/>\nConstruction: Join AC and BC.<br \/>\nProof: \u22201 = \u22202 = 90\u00b0 \u2026.[Tangent is I to the radius (through the point of contact<br \/>\n\u2234 APBC is a square.<br \/>\nLength of each tangent<br \/>\n= AP = PB = 4 cm<br \/>\n= AC = radius = 4 cm<\/p>\n<p>Also<\/p>\n<p><strong>Question 10. In the given figure, PQ and PR are two tangents to a circle with centre O. If \u2220QPR = 46\u00b0, then calculate \u2220QOR. (2014D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130428\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-15.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 15\" width=\"174\" height=\"119\" \/><br \/>\nSolution:<br \/>\n\u2220OQP = 900<br \/>\n\u2220ORP = 90\u00b0<br \/>\n\u2220OQP + \u2220QPR + \u2220ORP + \u2220QOR = 360\u00b0 \u2026[Angle sum property of a quad.<br \/>\n90\u00b0 + 46\u00b0 + 90\u00b0 + \u2220QOR = 360\u00b0<br \/>\n\u2220QOR = 360\u00b0 \u2013 90\u00b0 \u2013 46\u00b0 \u2013 90\u00b0 = 134\u00b0<\/p>\n<p><strong>Question 11. In the given figure, PA and PB are tangents to the circle with centre O such that \u2220APB = 50\u00b0. Write the measure of \u2220OAB. (2015D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130429\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-16.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 16Important Questions for Class 10 Maths Chapter 10 Circles 16\" width=\"192\" height=\"119\" \/><br \/>\nSolution:<br \/>\nPA = PB \u2026[\u2235 Tangents drawn from external point are equal<br \/>\n\u2220OAP = \u2220OBP = 90\u00b0<br \/>\n\u2220OAB = \u2220OBA \u2026 [Angles opposite equal sides<br \/>\n\u2220OAP + \u2220AOB + \u2220OBP + \u2220APB = 360\u00b0 \u2026 [Quadratic rule<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130430\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-17.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 17\" width=\"189\" height=\"118\" \/><br \/>\n90\u00b0 + \u2220AOB + 90\u00b0 + 50\u00b0 = 360\u00b0<br \/>\n\u2220AOB = 360\u00b0 \u2013 230\u00b0<br \/>\n= 130\u00b0<br \/>\n\u2220AOB + \u2220OAB + \u2220OBA = 180\u00b0 \u2026 [\u2206 rule<br \/>\n130\u00b0 + 2\u2220OAB = 180\u00b0 \u2026 [From (i)<br \/>\n2\u2220OAB = 50\u00b0<br \/>\n\u21d2 \u2220OAB = 25\u00b0<\/p>\n<p><strong>Question 12. From an external point P, tangents PA and PB are drawn to a circle with centre 0. If \u2220PAB = 50\u00b0, then find \u2220AOB. (2016D)<\/strong><br \/>\nSolution:<br \/>\nPA = PB \u2026[\u2235 Tangents drawn from external point are equal<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130431\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-18.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 18\" width=\"166\" height=\"110\" \/><br \/>\n\u2220PBA = \u2220PAB = 50\u00b0 \u2026[Angles equal to opposite sides<br \/>\nIn \u2206ABP, \u2220PBA + \u2220PAB + \u2220APB = 180\u00b0 \u2026[Angle-sum-property of a \u2206<br \/>\n50\u00b0 + 50\u00b0 + \u2220APB = 180\u00b0<br \/>\n\u2220APB = 180\u00b0 \u2013 50\u00b0 \u2013 50\u00b0 = 80\u00b0<br \/>\nIn cyclic quadrilateral OAPB<br \/>\n\u2220AOB + \u2220APB = 180\u00b0 \u2026\u2026[Sum of opposite angles of a cyclic (quadrilateral is 180\u00b0<br \/>\n\u2220AOB + 80o = 180\u00b0<br \/>\n\u2220AOB = 180\u00b0 \u2013 80\u00b0 = 100\u00b0<\/p>\n<p><strong>Question 13. In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If \u2220QPT = 60\u00b0, find \u2220PRQ. (2015OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130432\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-19.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 19\" width=\"141\" height=\"117\" \/><br \/>\nSolution:<br \/>\nPQ is the chord of the circle and PT is tangent.<br \/>\n\u2234 \u2220OPT = 90\u00b0 \u2026[Tangent is I to the radius through the point of contact<br \/>\nNow \u2220QPT = 60\u00b0 \u2026 [Given<br \/>\n\u2220OPQ = \u2220OPT \u2013 \u2220QPT<br \/>\n\u21d2 \u2220OPQ = 90\u00b0 \u2013 60\u00b0 = 30\u00b0<br \/>\nIn \u2206OPQ, OP = OQ<br \/>\n\u2220OQP = \u2220OPQ = 30\u00b0 \u2026 [In a \u2206, equal sides have equal \u2220s opp. them<br \/>\nNow, \u2220OQP + \u2220OPQ + \u2220POQ = 180\u00b0<br \/>\n\u2234 \u2220POQ = 120\u00b0 \u2026[\u2220POQ = 180o \u2013 (30\u00b0 + 30\u00b0)<br \/>\n\u21d2 Reflex \u2220POQ = 360\u00b0 \u2013 120\u00b0 = 240\u00b0 \u2026[We know that the angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle<br \/>\n\u2234 Reflex \u2220POQ = 2\u2220PRO<br \/>\n\u21d2 240\u00b0 = 2\u2220PRQ<br \/>\n\u21d2 \u2220PRQ = \\(\\frac{240^{\\circ}}{2}\\) = 120\u00b0<\/p>\n<p><strong>Question 14. In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). (2012D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130433\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-20.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 20\" width=\"142\" height=\"133\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130434\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-21.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 21\" width=\"140\" height=\"149\" \/><br \/>\nAP = AR = 4 cm<br \/>\nRC = 11 \u2013 4 = 7 cm<br \/>\nRC = QC = 7 cm<br \/>\nBQ = BP = 3 cm<br \/>\nBC = BQ + QC<br \/>\n= 3 + 7 = 10 cm<\/p>\n<p><strong>Question 15. In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. Calculate the radius of the circle inscribed in the triangle (in cm). (2014OD)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130435\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-22.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 22\" width=\"152\" height=\"187\" \/><br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> \u2026[Pythagoras\u2019 theorem<br \/>\n= (5)<sup>2<\/sup> + (12)<sup>2<\/sup><br \/>\nAC<sup>2<\/sup> = 25 + 144<br \/>\nAC = \\(\\sqrt{169}\\) = 13 cm<br \/>\nArea of \u2206ABC = Area of \u2206AOB + ar. of \u2206BOC + ar. of \u2206AOC<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130436\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-23.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 23\" width=\"391\" height=\"84\" \/><br \/>\n60 = r(AB + BC + AC)<br \/>\n60 = r(5 + 12 + 13)<br \/>\n60 = 30r \u21d2 r = 2 cm<\/p>\n<p><strong>Question 16. Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. (2011OD)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130437\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-24.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 24\" width=\"147\" height=\"139\" \/><br \/>\nRadius = R<br \/>\nAB = a + a = 2a<br \/>\n\u2234 Perimeter = 4(AB)<br \/>\n= 4(2a)<br \/>\n= 8a cm<\/p>\n<p><strong>Question 17. In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius DA of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD \u22a5 CD, then calculate the length of CD. (2013OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130438\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-25.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 25\" width=\"145\" height=\"130\" \/><br \/>\nSolution:<br \/>\nConst. Join OR<br \/>\nProof. \u22201 = \u22202 = 90\u00b0 \u2026 [Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u22203 = 90\u00b0 \u2026[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130439\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-26.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 26\" width=\"201\" height=\"161\" \/><br \/>\n\u2234 ORDS is a square.<br \/>\nDR = OS = 10 cm \u2026(i)<br \/>\nBP = BQ = 27 cm \u2026[Tangents drawn from an external point<br \/>\n\u2234 CQ = 38 \u2013 27 = 11 cm<br \/>\nRC = CO = 11 cm \u2026[Tangents drawn from an external point<br \/>\nDC = DR + RC = 10 + 11 = 21 cm \u2026[From (i) &amp; (ii)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Circles_Class_10_Important_Questions_Short_Answer-I_2_Marks\"><\/span>Circles Class 10 Important Questions Short Answer-I (2 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 18. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. (2012OD)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130440\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-27.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 27\" width=\"141\" height=\"131\" \/><br \/>\nProof: \u22201 = 90\u00b0 \u2026(i)<br \/>\n\u22202 = 90\u00b0 \u2026(ii)<br \/>\n\u22201 = \u22202 \u2026 [From (i) &amp; (ii)<br \/>\nBut these are alternate interior angles<br \/>\n\u2234PQ || RS<\/p>\n<p><strong>Question 19. In the figure, AB is the diameter of a circle with centre O and AT is a tangent. If \u2220AOQ = 58\u00b0, find \u2220ATQ.  (2015D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130441\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-28.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 28\" width=\"142\" height=\"127\" \/><br \/>\nSolution:<br \/>\n\u2220ABQ = \\(\\frac{1}{2}\\) \u2220AOQ = \\(\\frac{58^{\\circ}}{2}\\) = 29\u00b0<br \/>\n\u2220BAT = 90\u00b0 \u2026.[Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u2220ATQ = 180\u00b0 \u2013 (\u2220ABQ + \u2220BAT)<br \/>\n= 180 \u2013 (29 + 90) = 180\u00b0 \u2013 119\u00b0 = 61\u00b0<\/p>\n<p><strong>Question 20. Two concentric circles are of radii 7 cm and r cm respectively, where r &gt; 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r. (2011D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130442\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-29.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 29\" width=\"147\" height=\"140\" \/><br \/>\nGiven: OC = 7 cm, AB = 48 cm<br \/>\nTo find: r = ?<br \/>\n\u2220OCA = 90\u00b0 ..[Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u2234 OC \u22a5 AB<br \/>\nAC = \\(\\frac{1}{2}\\) (AB) \u2026 [\u22a5 from the centre bisects the chord<br \/>\n\u21d2 AC = \\(\\frac{1}{2}\\) (48) = 24 cm<br \/>\nIn rt. \u2206OCA, OA<sup>2<\/sup> = OC<sup>2<\/sup> + AC<sup>2<\/sup> \u2026 [Pythagoras\u2019 theorem<br \/>\nr<sup>2<\/sup> = (7)<sup>2<\/sup> + (24)<sup>2<\/sup><br \/>\n= 49 + 576 = 625<br \/>\n\u2234 r= \\(\\sqrt{625}\\) = 25 cm<\/p>\n<p><strong>Question 21. In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. (2012D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130443\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-30.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 30\" width=\"128\" height=\"129\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130444\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-31.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 31\" width=\"116\" height=\"116\" \/><br \/>\nConst.: Join OC<br \/>\nProof: AB is a tangent to smaller circle and OC is a radius.<br \/>\n\u2234 \u2220OCB = 90\u00b0 \u2026 above theorem<br \/>\nIn the larger circle, AB is a chord and OC \u22a5 AB.<br \/>\n\u2234 AC = CB \u2026 [\u22a5 from the centre bisects the chord<\/p>\n<p><strong>Question 22. In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD. (2011OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130446\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-32.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 32\" width=\"126\" height=\"150\" \/><br \/>\nSolution:<br \/>\nAB + CD = AD + BC<br \/>\n6 + 8 = AD + 9<br \/>\n14 \u2013 9 = AD \u21d2 AD = 5 cm<\/p>\n<p><strong>Question 23. Prove that the parallelogram circumscribing a circle is a rhombus. (2012D, 2013D)<\/strong><br \/>\nSolution:<br \/>\nGiven. ABCD is a \u0965<sup>gm<\/sup>.<br \/>\nTo prove. ABCD is a rhombus.<br \/>\nProof. In \u0965<sup>gm<\/sup>, opposite sides are equal<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130447\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-33.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 33\" width=\"141\" height=\"130\" \/><br \/>\nAB = CD<br \/>\nand AD = BC ..(i)<br \/>\nAP = AS \u2026[Tangents drawn from an external point are equal in length<br \/>\nPB = BQ<br \/>\nCR = CO<br \/>\nDR = DS<br \/>\nBy adding these tangents,<br \/>\n(AP + PB) + (CR + DR) = AS + BQ + CQ + DS<br \/>\nAB + CD = (AS + DS) + (BQ + CQ)<br \/>\nAB + CD = AD + BC<br \/>\nAB + AB = BC + BC \u2026 [From (i)<br \/>\n2AB = 2 BC<br \/>\nAB = BC \u2026(ii)<br \/>\nFrom (i) and (ii), AB = BC = CD = DA<br \/>\n\u2234 \u0965<sup>gm<\/sup> ABCD is a rhombus.<\/p>\n<p><strong>Question 24. In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. Prove that: AB + CD = BC + DA. (2013 OD, 2016 OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130448\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-34.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 34\" width=\"141\" height=\"138\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130449\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-35.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 35\" width=\"144\" height=\"133\" \/><br \/>\nAP = AS \u2026\u2026(i) (Tangents drawn from an external point are equal in length<br \/>\nBP = BO \u2026(ii)<br \/>\nCR = CQ \u2026.(iii)<br \/>\nDR = DS ..(iv)<br \/>\nBy adding (i) to (iv)<br \/>\n(AP + BP) + (CR + DR) = AS + BQ + CQ + DS<br \/>\nAB + CD = (BQ + CQ) + (AS + DS)<br \/>\n\u2234 AB + CD = BC + AD (Hence proved)<\/p>\n<p><strong>Question 25. In the given figure, an isosceles \u2206ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. (2012D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130450\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-36.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 36\" width=\"153\" height=\"133\" \/><br \/>\nSolution:<br \/>\nGiven: The incircle of \u2206ABC touches the sides BC, CA and AB at D, E and F F respectively.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130451\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-37.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 37\" width=\"158\" height=\"130\" \/><br \/>\nAB = AC<br \/>\nTo prove: BD = CD<br \/>\nProof: Since the lengths of tangents drawn from an external point to a circle are equal<br \/>\n\u2234 AF = AE \u2026 (i)<br \/>\nBF = BD \u2026(ii)<br \/>\nCD = CE \u2026(iii)<br \/>\nAdding (i), (ii) and (iii), we get<br \/>\nAF + BF + CD = AE + BD + CE<br \/>\n\u21d2 AB + CD = AC + BD<br \/>\nBut AB = AC \u2026 [Given<br \/>\n\u2234 CD = BD<\/p>\n<p><strong>Question 26. In Figure, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths 8 cm and 6 cm respectively, find the value of r. (2012OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130452\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-38.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 38\" width=\"137\" height=\"164\" \/><br \/>\nSolution:<br \/>\nConst.: Join AO, OB, CO<br \/>\nProof: area of \u2206ABC<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130453\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-39.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 39\" width=\"399\" height=\"377\" \/><br \/>\nFrom (i) and (ii), we get 12r = 24<br \/>\n\u2234 r = 2 cm<\/p>\n<p><strong>Question 27. In the given figure, a circle inscribed in \u2206ABC touches its sides AB, BC and AC at points D, E &amp; F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. (2013D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130454\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-40.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 40\" width=\"185\" height=\"155\" \/><br \/>\nSolution:<br \/>\nLet AD = AF = x<br \/>\nBD = BE = y \u2026[Two tangents drawn from and an external point are equal<br \/>\nCE = CF = z<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130455\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-41.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 41\" width=\"187\" height=\"154\" \/><br \/>\nAB = 12 cm \u2026[Given<br \/>\n\u2234 x + y = 12 cm \u2026(i)<br \/>\nSimilarly,<br \/>\ny + z = 8 cm \u2026(ii)<br \/>\nand x + z = 10 cm \u2026(iii)<br \/>\nBy adding (i), (ii) &amp; (iii)<br \/>\n2(x + y + z) = 30<br \/>\nx + y + z = 15 \u2026[\u2235 x + y = 12<br \/>\nz = 15 \u2013 12 = 3<br \/>\nPutting the value of z in (ii) &amp; (iii),<br \/>\ny + 3 = 8<br \/>\ny = 8 \u2013 3 = 5<br \/>\nx + 3 = 10<br \/>\nx = 10 \u2013 3 = 7<br \/>\n\u2234 AD = 7 cm, BE = 5 cm, CF = 3 cm<\/p>\n<p><strong>Question 28. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. (2014OD)<\/strong><br \/>\nSolution:<br \/>\nGiven: The incircle of \u2206ABC touches the sides BC, CA and AB at D, E and F respectively.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130456\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-42.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 42\" width=\"156\" height=\"130\" \/><br \/>\nAB = AC<br \/>\nTo prove: BD = CD<br \/>\nProof: AF = AE ..(i)<br \/>\nBF = BD \u2026(ii)<br \/>\nCD = CE \u2026(iii)<br \/>\nAdding (i), (ii) and (iii), we get<br \/>\nAF + BF + CD = AE + BD + CE<br \/>\n\u21d2 AB + CD = AC + BD<br \/>\nBut AB = AC \u2026[Given<br \/>\n\u2234 CD = BD<\/p>\n<p><strong>Question 29. In the figure, a \u2206ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. If the area of \u2206ABC is 54 cm<sup>2<\/sup>, then find the lengths of sides AB and AC. (2011D, 2011OD, 2015 OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130457\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-43.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 43\" width=\"205\" height=\"133\" \/><br \/>\nSolution:<br \/>\nGiven: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(\u2206ABC) = 54 cm<sup>2<\/sup><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130458\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-44.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 44\" width=\"204\" height=\"119\" \/><br \/>\nJoint: OA, OF, OE, OB and OC<br \/>\nLet AF = AE = x<br \/>\nBD = BF = 6 cm<br \/>\nCD = CE = 9 cm<br \/>\n\u2234 AB = AF + BF = x + 6 \u2026(i)<br \/>\nAC = AE + CE = x + 9 \u2026(ii)<br \/>\nBC = DB + CD = 6 + 9 = 15 cm \u2026(iii)<br \/>\nIn \u2206ABC,<br \/>\nArea of \u2206ABC = 54 cm<sup>2<\/sup> \u2026[Given<br \/>\nar(\u2206ABC) = ar(\u2206BOC) + ar(\u2206AOC) + ar(\u2206AOB)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130459\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-45.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 45\" width=\"397\" height=\"323\" \/><\/p>\n<p><strong>Question 30. In the figure, a circle is inscribed in a \u2206ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC, and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. (2016D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130460\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-46.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 46\" width=\"158\" height=\"133\" \/><br \/>\nSolution:<br \/>\nAB = 12 cm, BC = 8 cm, CA = 10 cm<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130461\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-47.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 47\" width=\"171\" height=\"139\" \/><br \/>\nAs we know,<br \/>\nAF = AD<br \/>\nCF = CE<br \/>\nBD = BE<br \/>\nLet AD = AF = x cm<br \/>\nthen, DB = AB \u2013 AD<br \/>\n= (12 \u2013 x) cm<br \/>\n\u2234 BE = (12 \u2013 x) cm ..[Tangents drawn from an external point are equal<br \/>\nSimilarly,<br \/>\nCF = CE = AC \u2013 AF = (10 \u2013 x) cm<br \/>\nBC = 8 cm \u2026[Given<br \/>\n\u21d2 BE + CE = 8 \u21d2 12 \u2013 x + 10 \u2013 x = 8<br \/>\n\u21d2 22 \u2013 8 = 2x \u21d2 2x = 14<br \/>\n\u2234 x = 7 \u2234 AD = x = 7 cm<br \/>\nBE = 12 \u2013 x = 12 \u2013 7 = 5 cm<br \/>\nCF = 10 \u2013 x = 10 \u2013 7 = 3 cm<\/p>\n<p><strong>Question 31. In Figure, common tangents AB and CD to the two circles with lo, centres O<sub>1<\/sub> and O<sub>2<\/sub> intersect at E. Prove that AB = CD. (2014OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130462\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-48.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 48\" width=\"177\" height=\"99\" \/><br \/>\nSolution:<br \/>\nEA = EC \u2026(i) \u2026.[Tangents drawn from an external point are equal<br \/>\nEB = ED \u2026(ii)<br \/>\nEA + EB = EC + ED \u2026[Adding (i) &amp; (ii)<br \/>\n\u2234 AB = CD (Hence proved)<\/p>\n<p><strong>Question 32. If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that \u2220QPR = 120\u00b0, prove that 2PQ = PO. (2014D)<\/strong><br \/>\nSolution:<br \/>\n\u2220OPQ = \\(\\frac{1}{2}\\)(\u2220QPR) ..[Tangents drawn from an external point are equal<br \/>\n= \\(\\frac{1}{2}\\)(120\u00b0) = 60\u00b0 \u2026[Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u2220OQP = 90\u00b0<br \/>\nIn rt. \u2206OQP, cos 60\u00b0 = \\(\\frac{PQ}{PO}\\)<br \/>\n\\(\\frac{1}{2}=\\frac{P Q}{P O}\\) \u2234 2PQ = PO<\/p>\n<p><strong>Question 33. From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQuestion (2015D)<\/strong><br \/>\nSolution:<br \/>\nIn \u2206s\u2019 TPC and TQC \u2026.[Tangents drawn from an external point are equal<br \/>\nTP = TQ<br \/>\nTC = TC \u2026[Common<br \/>\n\u22201 = \u22202 \u2026[TP and TQ are equally inclined to OT<br \/>\n\u2234 \u2206TPC = \u2206TQC \u2026 [SAS<br \/>\n\u2234 PC = QC \u2026[CPCT<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130463\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-49.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 49\" width=\"173\" height=\"124\" \/><br \/>\n\u22203 = \u22204 \u2026(i)<br \/>\n\u21d2 \u22203 + 24 = 180\u00b0 \u2026 [Linear pair<br \/>\n\u21d2 \u22203 + \u22203 = 180\u00b0\u2026[From (i)<br \/>\n\u21d2 2\u22203 = 180\u00b0 \u21d2 \u22203 = 90\u00b0<br \/>\n\u2234 \u22203 = \u22204 = 90\u00b0<br \/>\n\u2234 OT is the right bisector of PQuestion<\/p>\n<p><strong>Question 34. In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If \u2220PRQ = 120\u00b0, then prove that OR = PR + R. (2015OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130464\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-50.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 50\" width=\"131\" height=\"153\" \/><br \/>\nSolution:<br \/>\nJoin OP and OO<br \/>\n\u2220OPR = 90\u00b0<br \/>\nPR = RQ \u2026 [Tangents drawn from an external point are equal<br \/>\n\u2220PRO = \\(\\frac{1}{2}\\) \u2220PRQ = \\(\\frac{1}{2}\\) \u00d7 120\u00b0 = 60\u00b0<br \/>\nNow, In \u2206OPR,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130465\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-51.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 51\" width=\"134\" height=\"126\" \/><br \/>\n\u21d2 \u2220OPR + \u2220POR + \u2220ORP = 180\u00b0 \u2026[\u2206 Rule<br \/>\n\u21d2 90\u00b0 + \u2220POR + 60\u00b0 = 180\u00b0<br \/>\n\u21d2 \u2220POR + 150\u00b0 = 180\u00b0<br \/>\n\u21d2 \u2220POR = 30\u00b0<br \/>\n\u21d2 sin 30\u00b0 = \\(\\frac{PR}{OR}\\) \u21d2 \\(\\frac{1}{2}=\\frac{P R}{O R}\\)<br \/>\n\u21d2 OR = 2PR<br \/>\n\u21d2 OR = PR + QR (\u2235 PR = RQ) \u2026(Hence proved)<\/p>\n<p><strong>Question 35. In the figure, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and \u2220APB = 60\u00b0. Find the length of chord AB. (2016D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130466\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-52.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 52\" width=\"148\" height=\"162\" \/><br \/>\nSolution:<br \/>\nPA = PB \u2026[Tangents drawn from an external point are equal<br \/>\nGiven:<br \/>\n\u2220APB = 60\u00b0<br \/>\n\u2220PAB = \u2220PBA \u2026 (i) \u2026(Angles opposite to equal sides<br \/>\nIn \u2206PAB, \u2220PAB + \u2220PBA + \u2220APB = 180\u00b0 \u2026[Angle-sum-property of a \u2206<br \/>\n\u21d2 \u2220PAB + \u2220PAB + 60\u00b0 = 180\u00b0<br \/>\n\u21d2 2\u2220PAB = 180\u00b0 \u2013 60o = 120\u00b0<br \/>\n\u21d2 \u2220PAB = 60\u00b0<br \/>\n\u21d2 \u2220PAB = \u2220PBA = \u2220APB = 60\u00b0<br \/>\n\u2234 APAB is an equilateral triangle<br \/>\nHence, AB = AP = 5 cm \u2026[\u2235 All sides of an equilateral A are equal<\/p>\n<p><strong>Question 36. In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that \u2220OTS = \u2220OST = 30\u00b0. (2016OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130467\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-53.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 53\" width=\"194\" height=\"120\" \/><br \/>\nSolution:<br \/>\nLet \u2220TOP = \u03b8 \u2026[Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u2220OTP = 90\u00b0<br \/>\nOT = OS = r \u2026 [Given<br \/>\nIn rt. \u2206OTP, cos \u03b8 = \\(\\frac{\\mathrm{OT}}{\\mathrm{OP}}\\)<br \/>\n\u21d2 cos \u03b8 = \\(\\frac{r}{2 r}\\) \u21d2 cos \u03b8 = \\(\\frac{1}{2}\\)<br \/>\n\u21d2 cos \u03b8 = cos 60\u00b0 \u21d2 \u03b8 = 60\u00b0<br \/>\n\u2234 \u2220TOS = 60\u00b0 + 60\u00b0 = 120\u00b0<br \/>\nIn ATOS,<br \/>\n\u2220OTS = \u2220OST \u2026[Angles opposite to equal sides<br \/>\nIn \u2220TOS,<br \/>\n\u2220TOS + \u2220OTS + \u2220OST = 180\u00b0 \u2026 [Angle-sum-property of a \u2206<br \/>\n120\u00b0 + \u2220OTS + \u2220OTS = 180\u00b0 \u2026 [From (i)<br \/>\n2\u2220OTS = 180\u00b0 \u2013 120\u00b0<br \/>\n\u2220OTS = 60\u00b0\/2 = 30\u00b0<br \/>\n\u2234 \u2220OTS = \u2220OST = 30\u00b0<\/p>\n<p><strong>Question 37. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Question If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. (2017D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130468\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-54.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 54\" width=\"214\" height=\"105\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130469\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-55.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 55\" width=\"163\" height=\"106\" \/><br \/>\nPA = PB = 12 cm \u2026(i)<br \/>\nQC = AC = 3 cm \u2026(ii)<br \/>\nQD = BD = 3 cm \u2026(iii)<br \/>\nTo find: PC + PD<br \/>\n= (PA \u2013 AC) + (PB \u2013 BD)<br \/>\n= (12 \u2013 3) + (12 \u2013 3) \u2026 [From (i), (ii) &amp; (iii)<br \/>\n= 9 + 9 = 18 cm<\/p>\n<p><strong>Question 38. In the figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q. (2013 OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130470\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-56.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 56\" width=\"161\" height=\"112\" \/><br \/>\nSolution:<br \/>\nTo prove: PR = RQ<br \/>\nProof: PR = RC \u2026 (i)<br \/>\nQR = RC<br \/>\nFrom (i) and (ii), PR = QR (Hence proved)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Circles_Class_10_Important_Questions_Short_Answer-II_3_Marks\"><\/span>Circles Class 10 Important Questions Short Answer-II (3 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 39. In the figure, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the lengths of QM, RN and PI. (2012OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130471\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-57.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 57\" width=\"156\" height=\"195\" \/><br \/>\nSolution:<br \/>\nLet PL = PN = x cm<br \/>\nQL = QM = y cm<br \/>\nRN = MR = z cm<br \/>\nPQ = 10 cm = x + y = 10 \u2026(i)<br \/>\nQR = 8 cm = y + z = 8 \u2026(ii)<br \/>\nPR = 12 cm = x + z = 12 \u2026(iii)<br \/>\nBy adding (i), (ii) and (iii),<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130472\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-58.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 58\" width=\"166\" height=\"163\" \/><br \/>\nWe get,<br \/>\n\u21d2 2x + 2y + 2z = 10 + 8 + 12<br \/>\n\u21d2 2(x + y + z) = 30<br \/>\n\u21d2 x + y + z = 15<br \/>\n\u21d2 10 + z = 15 \u2026 [From (i)<br \/>\n\u2234 z = 15 \u2013 10 = 5 cm<br \/>\nFrom (ii)<br \/>\ny + 5 = 8<br \/>\ny = 8 \u2013 5<br \/>\ny = 3 cm<br \/>\nFrom (iii)<br \/>\nx + 5 = 12<br \/>\nx = 12 \u2013 5<br \/>\nx = 7 cm<br \/>\n\u2234 QM = 3 cm, RN = 5 cm, PL = 7 cm<\/p>\n<p><strong>Question 40. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (2012D)<\/strong><br \/>\nSolution:<br \/>\n1st method:<br \/>\nTo prove. (i) \u2220AOD + \u2220BOC = 180\u00b0<br \/>\n(ii) \u2220AOB + \u2220COD = 180\u00b0<br \/>\nProof. In \u2206BPO and \u2206BQO \u2026[Tangents drawn from an external point are equal<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130473\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-59.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 59\" width=\"138\" height=\"128\" \/><br \/>\nPO = 20 \u2026 [radii<br \/>\nBO = BO \u2026 [Common<br \/>\n\u2206BPO = \u2206BQO \u2026 [SSS Congruency rule<br \/>\n\u22208 = \u22201 \u2026(i) (c.p.c.t.)<br \/>\nSimilarly,<br \/>\n\u22202 = \u22203, \u22204 = \u22205 and \u22206 = \u22207<br \/>\n\u22201 + \u22202 + 23+ 24 + 25 + 26+ 27 + \u22208 = 360\u00b0 \u2026(Complete angles<br \/>\n\u22201 + \u22202 + 22+ 25 + 25 + 26+ \u22206+ \u22201 = 360\u00b0<br \/>\n2(\u22201 + \u22202 + 25 + 26) = 360\u00b0<br \/>\n\u2220BOC + \u2220AOD = 180\u00b0\u2026(i) [Proved part I<br \/>\n\u2220AOB + \u2220BOC + \u2220COD + \u2220DOA = 360\u00b0 \u2026(Complete angles<br \/>\n\u2220AOB + \u2220COD + 180o = 360\u00b0 \u2026 [From (i)<br \/>\n\u2234 \u2220AOB + \u2220COD = 360\u00b0 \u2013 180o = 180\u00b0 \u2026(proved)<\/p>\n<p>2nd method:<br \/>\nTo prove:<br \/>\n(i) \u22206 + \u22208 = 180\u00b0<br \/>\n(ii) \u22205 + \u22207 = 180\u00b0<br \/>\nProof. As AS and AP are tangents to the circle from a point A<br \/>\n\u2234 O lies on the bisector of \u2220SAP<br \/>\n\u2234 \u22201 = \\(\\frac{1}{2}\\) \u2220BAD \u2026(i)<br \/>\nSimilarly BO, CO and DO are the bisectors of<br \/>\n\u2220ABC, \u2220BCD and \u2220ADC respectively. \u2026(ii)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130474\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-60.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 60\" width=\"398\" height=\"220\" \/><br \/>\n\u2234 \u22201 + \u22204 + \u22203 + \u22202 =180\u00b0\u2026(iii) ..[From (1) &amp; (ii)<br \/>\nIn \u2206AOD, \u22201 + \u22202 + 26 = 180\u00b0 \u2026[Angle-sum-Prop. of a \u2206<br \/>\nIn \u2206BOC, \u22203 + \u22204 + \u22208 = 180\u00b0 \u2026(v)<br \/>\nAdding (iv) and (v)<br \/>\n(\u22201 + \u22202 + 23 + 24) + 26 + 28 = 180\u00b0 + 180\u00b0<br \/>\n180\u00b0 + 26 + 28 = 180\u00b0 + 180\u00b0 \u2026 [From (iii)<br \/>\n\u2234\u22206 + 28 = 180\u00b0<br \/>\nNow \u22205 + \u22206 + \u22207 + \u22208 = 360\u00b0 \u2026 (Complete angles<br \/>\n(\u22205 + \u22207) + (\u22206 + \u22208) = 360\u00b0<br \/>\n(\u22205 + \u22207) + 180\u00b0 = 360\u00b0<br \/>\n\u22205 + \u22207 = 360\u00b0 \u2013 180\u00b0 = 180\u00b0<br \/>\n\u22205 + \u22207 = 180\u00b0<\/p>\n<p><strong>Question 41. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that \u2220PTQ = 2\u2220OPQ. (2017D)<\/strong><br \/>\nSolution:<br \/>\nWe are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Figure).<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130475\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-61.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 61\" width=\"187\" height=\"129\" \/><br \/>\nWe need to prove that:<br \/>\n\u2220PTQ = 2\u2220OPQ<br \/>\nLet \u2220PTQ = \u03b8<br \/>\nNow, TP = TQuestion \u2026.[\u2235 Lengths of tangents drawn from an external pt. to a circle are equal<br \/>\nSo, TPQ is an isosceles triangle.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130476\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-62.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 62\" width=\"408\" height=\"229\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Circles_Class_10_Important_Questions_Long_Answer_4_Marks\"><\/span>Circles Class 10 Important Questions Long Answer (4 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 42. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (2011OD, 2012OD, 2013D, 2014OD, 2015D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130477\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-63.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 63\" width=\"168\" height=\"146\" \/><br \/>\nGiven: XY is a tangent at point P to the circle with centre O.<br \/>\nTo prove: OP \u22a5 XY<br \/>\nConst.: Take a point Q on XY other than P and join to OQuestion<br \/>\nProof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.<br \/>\n\u2234 OQ &gt; OP<br \/>\nThis happens with every point on the line XY except the point P.<br \/>\nOP is the shortest of all the distances of the point O to the points of XY<br \/>\n\u2234 OP \u22a5 XY \u2026 [Shortest side is \u22a5<\/p>\n<p>Question 43.<br \/>\nProve that the lengths of tangents drawn from an external point to a circle are equal. (2011D, 2012OD, 2013OD, 2014, 2015D &amp; OD<br \/>\n2016D &amp; OD, 2017D)<br \/>\nSolution:<br \/>\nGiven: PT and PS are tangents from an external point P to the circle with centre O.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130478\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-64.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 64\" width=\"218\" height=\"131\" \/><br \/>\nTo prove: PT = PS<br \/>\nConst.: Join O to P,<br \/>\nT &amp; S<br \/>\nProof: In \u2206OTP and<br \/>\n\u2206OSP,<br \/>\nOT = OS \u2026[radii of same circle<br \/>\nOP = OP \u2026[circle<br \/>\n\u2220OTP \u2013 \u2220OSP \u2026[Each 90\u00b0<br \/>\n\u2234 AOTP = AOSP \u2026[R.H.S<br \/>\nPT = PS \u2026[c.p.c.t<\/p>\n<p>Question 44.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130479\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-65.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 65\" width=\"226\" height=\"126\" \/><br \/>\nIn the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP. (2014OD)<br \/>\nSolution:<br \/>\nTP = TQuestion .. [Tangents drawn from an external point<br \/>\n\u2206TPQ is an isosceles \u2206 and TO is the bisector of \u2220PTQ ,<br \/>\nOT \u22a5 PQ \u2026[Tangent is \u22a5 to the radius through the point of contact<br \/>\n\u2234 OT bisects PQ<br \/>\n\u2234 PR = RQ = 16 = 8 cm \u2026[Given<br \/>\nIn rt. \u2206PRO,<br \/>\nPR<sup>2<\/sup> + RO<sup>2<\/sup> = PO<sup>2<\/sup> \u2026 [Pythagoras\u2019 theorem<br \/>\n8<sup>2<\/sup> + RO<sup>2<\/sup> = (10)<sup>2<\/sup><br \/>\nRO<sup>2<\/sup> = 100 \u2013 64 = 36<br \/>\n\u2234 RO = 6 cm<br \/>\nLet TP = x cm and TR = y cm<br \/>\nThen OT = (y + 6) cm<br \/>\nIn rt. \u2206PRT, x<sup>2<\/sup> = y<sup>2<\/sup> + 8<sup>2<\/sup> \u2026(i) \u2026[Pythagoras\u2019 theorem<br \/>\nIn rt. \u2206OPT,<br \/>\nOT<sup>2<\/sup> = TP<sup>2<\/sup> + PO<sup>2<\/sup> \u2026(Pythagoras\u2019 theorem<br \/>\n(y + 6)<sup>2<\/sup> = x<sup>2<\/sup> + 10<sup>2<\/sup><br \/>\ny<sup>2<\/sup> + 12y + 36 = y<sup>2<\/sup> +64 + 100 \u2026[From (i)<br \/>\n12y = 164 \u2013 36 = 128 \u21d2 y = \\(\\frac{128}{12}=\\frac{32}{3}\\)<br \/>\nPutting the value of y in (i),<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130480\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-66.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 66\" width=\"294\" height=\"122\" \/><\/p>\n<p><strong>Question 45. In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that \u2220RPQ = 30\u00b0. A chord RS is drawn parallel to the tangent PQuestion Find \u2220RQS. (2015D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130481\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-67.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 67\" width=\"140\" height=\"98\" \/><br \/>\nSolution:<br \/>\nPR = PO \u2026[\u2235 Tangents drawn from an external point are equal<br \/>\n\u21d2 \u2220PRQ = \u2220PQR \u2026[\u2235 Angles opposite equal sides are equal<br \/>\nIn \u2206PQR,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130482\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-68.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 68\" width=\"178\" height=\"120\" \/><br \/>\n\u21d2 \u2220PRQ + \u2220RPQ + \u2220POR = 180\u00b0\u2026[\u2206 Rule<br \/>\n\u21d2 30\u00b0 + 2\u2220PQR = 180\u00b0<br \/>\n\u21d2 \\(\\angle \\mathrm{PQR}=\\frac{(180-30)^{\\circ}}{2}\\)<br \/>\n= 75\u00b0<br \/>\n\u21d2 SR || QP and QR is a transversal<br \/>\n\u2235 \u2220SRQ = \u2220PQR \u2026 [Alternate interior angle<br \/>\n\u2234 \u2220SRO = 75\u00b0 \u2026..[Tangent is I to the radius through the point of contact<br \/>\n\u21d2 \u2220ORP = 90\u00b0<br \/>\n\u2234 \u2220ORP = \u2220ORQ + \u2220QRP<br \/>\n90\u00b0 = \u2220ORQ + 75\u00b0<br \/>\n\u2220ORQ = 90\u00b0 \u2013 75o = 150<br \/>\nSimilarly, \u2220RQO = 15\u00b0<br \/>\nIn \u2206QOR,<br \/>\n\u2220QOR + \u2220QRO + \u2220OQR = 180\u00b0 \u2026[\u2206 Rule<br \/>\n\u2234\u2220QOR + 15\u00b0 + 15\u00b0 = 180\u00b0<br \/>\n\u2220QOR = 180\u00b0 \u2013 30\u00b0 = 150\u00b0<br \/>\n\u21d2 \u2220QSR = \\(\\frac{1}{2}\\)\u2220QOR<br \/>\n\u21d2 \u2220QSR = \\(\\frac{150^{\\circ}}{2}\\) = 750 \u2026 [Used \u2220SRQ = 75\u00b0 as solved above<br \/>\nIn ARSQ, \u2220RSQ + \u2220QRS + \u2220RQS = 180\u00b0 \u2026 [\u2206 Rule<br \/>\n\u2234 75\u00b0 + 75\u00b0 + \u2220RQS = 180\u00b0<br \/>\n\u2220RQS = 180\u00b0 \u2013 150o = 30\u00b0<\/p>\n<p><strong>Question 46. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. (2015OD)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130483\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-69.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 69\" width=\"141\" height=\"123\" \/><br \/>\nB is the mid point of arc (ABC)<br \/>\nOA = OC \u2026[Radius<br \/>\nOF = OF \u2026[Common<br \/>\n\u2234 \u22201 = \u22202 \u2026[Equal angles opposite equal sides<br \/>\n\u2234 \u2206OAF = \u2206OCF (SAS)<br \/>\n\u2234 \u2220AFO = \u2220CFO = 90\u00b0 \u2026[c.p.c.t<br \/>\n\u21d2 \u2220AFO = \u2220DBO = 90\u00b0 \u2026[Tangent is \u22a5to the radius through the point of contact<br \/>\nBut these are corresponding angles,<br \/>\n\u2234 AC || DE<\/p>\n<p><strong>Question 47. In the figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. (2016D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130484\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-70.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 70\" width=\"200\" height=\"125\" \/><br \/>\nSolution:<br \/>\n\u2220OPT = 90\u00b0 \u2026[Tangent is \u22a5 to the radius through the point of contact<br \/>\nWe have, OP = 5 cm, OT = 13 cm<br \/>\nIn rt. \u2206OPT,<br \/>\nOP<sup>2<\/sup> + PT<sup>2<\/sup> = OT? \u2026[Pythagoras\u2019 theorem<br \/>\n\u21d2 (5)<sup>2<\/sup> + PT<sup>2<\/sup> = (13)<sup>2<\/sup><br \/>\n\u21d2 PT<sup>2<\/sup> = 169 \u2013 25 = 144 cm<br \/>\n\u21d2 PT = \\(\\sqrt{144}\\)<br \/>\n= 12 cm<br \/>\nOP = OQ = OE = 5 cm \u2026 [Radius of the circle<br \/>\nET = OT \u2013 OE<br \/>\n= 13 \u2013 5 = 8 cm<br \/>\nLet, PA = x cm, then AT = (12 \u2013 x) cm<br \/>\nPA = AE = x cm \u2026[Tangent drawn from an external point<br \/>\nIn rt. \u2206AET,<br \/>\nAE<sup>2<\/sup> + ET<sup>2<\/sup> = AT<sup>2<\/sup> \u2026(Pythagoras\u2019 theorem<br \/>\n\u21d2 x<sup>2<\/sup> + (8)<sup>2<\/sup> = (12 \u2013 x)<sup>2<\/sup><br \/>\n\u21d2 x<sup>2<\/sup> + 64 = 144 + x<sup>2<\/sup> \u2013 24x<br \/>\n\u21d2 24x = 144 \u2013 64<br \/>\nx = \\(\\frac{80}{24}=\\frac{10}{3}\\) cm<br \/>\nAB = AE + EB = AE + AE = 2AE = 2x :<br \/>\n\u2234 AB = \\(2\\left(\\frac{10}{3}\\right)=\\frac{20}{3} \\mathrm{cm}=6 \\frac{2}{3}\\) cm<br \/>\nor 6.67 cm or 6.6 cm<\/p>\n<p><strong>Question 48. In the figure, two equal circles, with centres 0 and O\u2019, touch each other A at X. OO\u2019 produced meets the circle with centre O\u2019 at A. AC is tangent to the circle with centre O, at the point C. O\u2019D is perpendicular to AC. Find the value of \\(\\frac{\\mathrm{DO}^{\\prime}}{\\mathrm{CO}}\\). (2016OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130485\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-71.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 71\" width=\"168\" height=\"102\" \/><br \/>\nSolution:<br \/>\nGiven: Two equal circles, with centres O and O\u2019, touch each other at point X. OO\u2019 is produced to meet the circle with centre (at A. AC is tangent to the circle with centre O, at the point C. O\u02bbD is perpendicular to AC.<br \/>\nTo find: = \\(\\frac{\\mathrm{DO}^{\\prime}}{\\mathrm{CO}}\\)<br \/>\nProof: \u2220ACO = 90\u00b0 \u2026 [Tangent is \u22a5 to the radius through the point of contact<br \/>\nIn \u2206AO\u2019D and \u2206AOC<br \/>\n\u2220O\u2019AD = \u2220OAC \u2026(Common<br \/>\n\u2234 \u2220ADO = \u2220ACO \u2026[Each 90\u00b0<br \/>\n\u2234 \u2206AO\u2019D ~ \u2234AOC \u2026(AA similarity<br \/>\n\\(\\frac{\\mathrm{AO}^{\\prime}}{\\mathrm{AO}}=\\frac{\\mathrm{DO}^{\\prime}}{\\mathrm{CO}}\\) \u2026 [In ~ As corresponding sides are proportional<br \/>\n\\(\\frac{r}{3 r}=\\frac{\\mathrm{DO}^{\\prime}}{\\mathrm{CO}}\\) \u2026[Let AO\u2019 = O\u2019X = OX = r \u21d2 AO = r +r+ r = 3r<br \/>\n\u2234 \\(\\frac{\\mathrm{DO}^{\\prime}}{\\mathrm{CO}}=\\frac{1}{3}\\)<\/p>\n<p><strong>Question 49. In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that \u2220DOE = 90\u00b0. (2013D)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130486\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-72.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 72\" width=\"178\" height=\"123\" \/><br \/>\nSolution:<br \/>\nProof: Let I be XY and m be XY\u2019<br \/>\n\u2220XDE + \u2220X\u2019ED = 180\u00b0 \u2026 [Consecutive interior angles<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130487\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-73.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 73\" width=\"179\" height=\"134\" \/><br \/>\n\\(\\frac{1}{2}\\)XDE + \\(\\frac{1}{2}\\)\u2220X\u2019ED =<br \/>\n= \\(\\frac{1}{2}\\) (180\u00b0)<br \/>\n= \u22201 + \u22202 = 90\u00b0 \u2026[OD is equally inclined to the tangents<br \/>\nIn \u2206DOE, \u22201 + \u22202 + 23 = 180\u00b0 \u2026[Angle-sum-property of a \u2206<br \/>\n90\u00b0 + 23 = 180\u00b0<br \/>\n\u21d2 \u22203 = 180\u00b0 \u2013 90o = 90\u00b0<br \/>\n\u2234 \u2220DOE = 90\u00b0 \u2026(proved)<\/p>\n<p><strong>Question 50. In the figure, the sides AB, BC and CA of triangle ABC touch a circle with centre o and radius r at P, Q and R. respectively. (2013OD)<\/strong><br \/>\n<strong>Prove that:<\/strong><br \/>\n<strong>(i) AB + CQ = AC + BQ<\/strong><br \/>\n<strong>(ii) Area (AABC) = \\(\\frac{1}{2}\\) (Perimeter of \u2206ABC ) \u00d7 r<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130488\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-74.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 74\" width=\"144\" height=\"153\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130489\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-75.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 75\" width=\"163\" height=\"165\" \/><br \/>\nPart I:<br \/>\nProof: AP = AR \u2026(i)<br \/>\nBP = BQ \u2026 (ii)<br \/>\nCQ = CR \u2026 (iii)<br \/>\nAdding (i), (ii) &amp; (iii)<br \/>\nAP + BP + CQ<br \/>\n= AR + BQ + CR<br \/>\nAB + CQ = AC + BQ<br \/>\nPart II: Join OP, OR, OQ, OA, OB and OC<br \/>\nProof: OQ \u22a5 BC; OR \u22a5 AC; OP \u22a5 AB<br \/>\nar(\u2206ABC) = ar(\u2206AOB) + ar(\u2206BOC) + ar (\u2206AOC)<br \/>\nArea of (\u2206ABC)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130490\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-76.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 76\" width=\"399\" height=\"230\" \/><\/p>\n<p><strong>Question 51. In the figure, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC. (2014OD)<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130491\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-77.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 77\" width=\"173\" height=\"144\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130492\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-78.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 78\" width=\"161\" height=\"142\" \/><br \/>\nLet AE = x \u2234 AF = x<br \/>\nBC = 8 + 6 = 14 cm<br \/>\nAB = (x + 8) cm<br \/>\nAC = (x + 6) cm<br \/>\n\u22201 = \u22202 = 23 = 90\u00b0 \u2026[ Tangent is \u22a5 to the radius B [through the point of contact<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130493\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-79.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 79\" width=\"397\" height=\"416\" \/><br \/>\n\\(4 \\sqrt{3 x(x+14)}\\) = 2(2x + 28)<br \/>\n\\(4 \\sqrt{3 x(x+14)}\\) = 2.2(x + 14)<br \/>\n3x(x + 14) = (x + 14)<sup>2<\/sup> \u2026 [Squaring both sides<br \/>\n3x(x + 14) \u2013 (x + 14)<sup>2<\/sup> = 0<br \/>\n(x + 14) [3x \u2013 (x + 14)] = 0<br \/>\n(x + 14) (2x \u2013 14) = 0<br \/>\nx = -14 or x = 7<br \/>\n\u2234 x = 7 \u2026 [As side of \u2206 cannot be -ve<br \/>\n\u2234 AB = x + 8 = 15 cm<br \/>\nand AC = x + 6 = 13 cm<\/p>\n<p><strong>Question 52. rove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre. (2014 D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130494\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-10-Circles-80.png\" alt=\"Important Questions for Class 10 Maths Chapter 10 Circles 80\" width=\"147\" height=\"145\" \/><br \/>\nGiven: CD and EF are two C parallel tangents at points A and B of a circle with centre O.<br \/>\nTo prove: AB passes through centre O or AOB is diameter of the circle.<br \/>\nConst.: Join OA and OB. Draw OM || CD.<br \/>\nProof: \u22201 = 90\u00b0 \u2026 (i)<br \/>\n\u2026[\u2235 Tangent is I to the radius through the point of contact<br \/>\nOM || CD<br \/>\n\u2234 \u22201 + \u22202 = 180\u00b0 \u2026(Co-interior angles<br \/>\n90\u00b0 + \u22202 = 180\u00b0 \u2026[From (i)<br \/>\n\u22202 = 180\u00b0 \u2013 90o = 90\u00b0<br \/>\nSimilarly, \u22203 = 90\u00b0<br \/>\n\u22202 + \u22203 = 90\u00b0 + 90\u00b0 = 180\u00b0<br \/>\n\u2234 AOB is a straight line.<br \/>\nHence AOB is a diameter of the circle with centre O.<br \/>\n\u2234 AB passes through centre 0.<\/p>\n<h4><\/h4>\n","protected":false},"excerpt":{"rendered":"<p>In Chapter 10 of Class 10 Maths, titled &#8220;Circles,&#8221; several important concepts are explored. 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