NCERT Solutions For Class 10 Maths Chapter 10 Circles 2026-27

NCERT Solutions for Class 10 Maths Chapter 10 Circles 2026-27 help students understand every important concept of circles in a simple and clear way. This chapter mainly explains tangents to a circle and the properties related to them. A tangent is a line that touches a circle at only one point. Students also learn that the tangent at any point of a circle is perpendicular to the radius passing through the point of contact. Another key result is that the lengths of two tangents drawn from an external point to a circle are equal.

The NCERT Solutions for Class 10 Maths Chapter 10 Circles 2026-27 provide easy, step-by-step answers to all the questions given in the latest NCERT textbook. These solutions are useful for understanding the method used in each problem instead of simply memorising the final answer. Clear explanations, correct formulas, and properly drawn figures make the chapter easier to study.

Infinity Learn offers these solutions to support students during homework, revision, and exam preparation. The answers are written in simple words so that learners can follow each step without confusion. By practising the textbook questions with these solutions, students can improve their understanding of geometrical proofs and learn how to present answers correctly in the board examination.

This chapter is important because questions based on tangents, radii, and circle properties are commonly asked in exams. Regular practice can help students identify the correct theorem and apply it confidently. The solutions can also be used to check answers and correct mistakes. They are helpful for quick revision before school tests, unit tests, and board exams.

Download NCERT Solutions For Class 10 Maths Chapter 10 Circles PDF

 Students can download the NCERT Solutions for Class 10 Maths Chapter 10 Circles PDF from Infinity Learn for easy and convenient study. The PDF includes clear, step-by-step solutions to all the questions given in the NCERT textbook. It explains important concepts such as tangents, points of contact, radii, and the properties of tangent.

Access NCERT Solutions for Class 10 Maths Chapter 10 Circles

Exercise 10.1

Question 1: How many tangents can a circle have?

Solution:

A circle contains infinitely many points on its circumference. A tangent can be drawn at every point on the circumference.

Therefore, a circle can have infinitely many tangents.

Question 2: Fill in the blanks

Question 3: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at Q. If OQ = 12 cm, find PQ.

Options: (A) 12 cm, (B) 13 cm, (C) 8.5 cm, (D) √119 cm

Solution:

The radius drawn to the point of contact is perpendicular to the tangent. Therefore:

OP is perpendicular to PQ.

Thus, triangle OPQ is a right-angled triangle in which:

• OP = 5 cm
• OQ = 12 cm
• PQ is the required tangent length

Using the Pythagoras theorem:

OQ² = OP² + PQ²

PQ² = OQ² - OP²

PQ² = 12² - 5²

PQ² = 144 - 25

PQ² = 119

PQ = √119 cm

Therefore, the correct answer is option (D), √119 cm.

Question 4: Draw a circle and two lines parallel to a given line such that one line is a tangent and the other is a secant.

Solution:

1. Draw a circle with centre O.
2. Draw the given line and name it l.
3. Draw a radius OT perpendicular to line l, where T lies on the circle.
4. Through T, draw a line parallel to l. Since this line is perpendicular to radius OT at T, it is a tangent to the circle.
5. Draw another line parallel to l at a distance from O that is less than the radius of the circle.
6. This second line will intersect the circle at two points, so it is a secant.

Hence, one of the parallel lines touches the circle at exactly one point, while the other cuts the circle at two points.

Exercise 10.2

Question 1: From a point Q, the length of a tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.

Options: (A) 7 cm, (B) 12 cm, (C) 15 cm, (D) 24.5 cm

Solution:

Let O be the centre of the circle and let PQ be the tangent drawn from Q. Then:

• PQ = 24 cm
• OQ = 25 cm
• OP is the radius of the circle

The radius OP is perpendicular to tangent PQ at P. Therefore, triangle OPQ is right-angled at P.

Using the Pythagoras theorem:

OQ² = OP² + PQ²

OP² = OQ² - PQ²

OP² = 25² - 24²

OP² = 625 - 576

OP² = 49

OP = √49 = 7 cm

Therefore, the radius of the circle is 7 cm. The correct answer is option (A).

Question 2: TP and TQ are tangents to a circle with centre O. If ∠POQ = 110°, find ∠PTQ.

Options: (A) 60°, (B) 70°, (C) 80°, (D) 90°

Solution:

Since TP and TQ are tangents, the radii drawn to the points of contact are perpendicular to them.

Therefore:

∠OPT = 90°

∠OQT = 90°

Now consider quadrilateral OPTQ. The sum of its interior angles is 360°.

∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°

90° + 90° + 110° + ∠PTQ = 360°

290° + ∠PTQ = 360°

∠PTQ = 360° - 290°

∠PTQ = 70°

Therefore, the correct answer is option (B), 70°.

Question 3: Tangents PA and PB are drawn from an external point P to a circle with centre O. If the angle between the tangents is 80°, find ∠POA.

Options: (A) 50°, (B) 60°, (C) 70°, (D) 80°

Solution:

The radius at a point of contact is perpendicular to the tangent. Hence:

∠OAP = 90°

∠OBP = 90°

Also, the angle between tangents PA and PB is:

∠APB = 80°

In quadrilateral OAPB, the sum of the four interior angles is 360°.

∠AOB + ∠OAP + ∠OBP + ∠APB = 360°

∠AOB + 90° + 90° + 80° = 360°

∠AOB + 260° = 360°

∠AOB = 100°

Now compare triangles OAP and OBP:

• OA = OB, because they are radii of the same circle.
• PA = PB, because tangents drawn from the same external point are equal.
• OP is common to both triangles.

Therefore, triangles OAP and OBP are congruent by the SSS rule.

Hence, OP bisects ∠AOB.

∠POA = (1)/(2) × ∠AOB

∠POA = (1)/(2) × 100°

∠POA = 50°

Therefore, the correct answer is option (A), 50°.

Question 4: Prove that the tangents drawn at the endpoints of a diameter of a circle are parallel.

Proof:

Let P and Q be the endpoints of a diameter PQ of a circle with centre O. Draw tangents at P and Q.

A tangent to a circle is perpendicular to the radius through the point of contact.

Therefore:

• The tangent at P is perpendicular to OP.
• The tangent at Q is perpendicular to OQ.

Since OP and OQ are parts of the same straight line PQ, both tangents are perpendicular to the same line PQ.

Two lines that are perpendicular to the same line are parallel to each other.

Therefore, the tangents at the endpoints of a diameter are parallel. Hence proved.

Question 5: Prove that the perpendicular drawn at the point of contact to the tangent of a circle passes through the centre.

Proof:

Let AB be a tangent to a circle at point P, and let O be the centre of the circle.

According to the tangent-radius theorem, the radius drawn to the point of contact is perpendicular to the tangent.

Therefore:

OP is perpendicular to AB.

Through a given point on a line, only one line can be drawn perpendicular to that line.

Thus, the perpendicular drawn at P to tangent AB must be the line OP itself.

Since OP contains the centre O, the perpendicular at the point of contact passes through the centre of the circle. Hence proved.

Question 6: The length of a tangent from a point A is 4 cm, and the distance of A from the centre of the circle is 5 cm. Find the radius.

Solution:

Let O be the centre of the circle and AB be the tangent from A, touching the circle at B.

Given:

• OA = 5 cm
• AB = 4 cm
• OB = radius of the circle

Radius OB is perpendicular to tangent AB. Therefore, triangle OBA is right-angled at B.

Using the Pythagoras theorem:

OA² = OB² + AB²

OB² = OA² - AB²

OB² = 5² - 4²

OB² = 25 - 16

OB² = 9

OB = √9 = 3 cm

Therefore, the radius of the circle is 3 cm.

Question 7: Two concentric circles have radii 5 cm and 3 cm. Find the length of a chord of the larger circle that touches the smaller circle.

Solution:

Let O be the common centre of the two circles. Let PQ be a chord of the larger circle that touches the smaller circle at A.

Since PQ is tangent to the smaller circle at A, the radius OA is perpendicular to PQ.

Given:

• OP = 5 cm, the radius of the larger circle
• OA = 3 cm, the radius of the smaller circle

In right-angled triangle OAP:

OP² = OA² + AP²

AP² = OP² - OA²

AP² = 5² - 3²

AP² = 25 - 9

AP² = 16

AP = 4 cm

The perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore:

AP = AQ = 4 cm

PQ = AP + AQ

PQ = 4 + 4

PQ = 8 cm

Therefore, the required length of the chord is 8 cm.

Question 8: A quadrilateral ABCD circumscribes a circle. Prove that AB + CD = AD + BC.

Proof:

Suppose the circle touches sides AB, BC, CD and DA at P, Q, R and S respectively.

Tangents drawn from the same external point have equal lengths. Therefore:

• AP = AS
• BP = BQ
• CQ = CR
• DR = DS

Now write each side as the sum of its two tangent segments:

AB = AP + PB

BC = BQ + QC

CD = CR + RD

AD = AS + SD

Consider the sum AB + CD:

AB + CD = AP + PB + CR + RD

Using AP = AS, PB = BQ, CR = CQ and RD = SD:

AB + CD = AS + BQ + CQ + SD

AB + CD = (AS + SD) + (BQ + QC)

AB + CD = AD + BC

Therefore, AB + CD = AD + BC. Hence proved.

Question 9: XY and X'Y' are two parallel tangents to a circle with centre O. Another tangent AB touches the circle at C and intersects XY at A and X'Y' at B. Prove that ∠AOB = 90°.

Proof:

Let XY touch the circle at P and let X'Y' touch it at Q. The third tangent AB touches the circle at C.

From the external point A, the two tangent segments are equal:

AP = AC

Also:

• OP = OC, because both are radii.
• OA is common to triangles OPA and OCA.

Therefore, triangles OPA and OCA are congruent by the SSS rule.

Hence, OA bisects the angle between the tangents AP and AC.

Thus:

∠OAB = (1)/(2) ∠PAC

Similarly, from external point B:

BQ = BC

Using OQ = OC and common side OB, triangles OQB and OCB are congruent.

Therefore, OB bisects the angle between BQ and BC.

Thus:

∠ABO = (1)/(2) ∠QBC

Since XY is parallel to X'Y' and AB is a transversal, the interior angles on the same side are supplementary:

∠PAC + ∠QBC = 180°

Therefore:

∠OAB + ∠ABO = (1)/(2)(∠PAC + ∠QBC)

∠OAB + ∠ABO = (1)/(2) × 180°

∠OAB + ∠ABO = 90°

In triangle AOB:

∠AOB + ∠OAB + ∠ABO = 180°

∠AOB + 90° = 180°

∠AOB = 90°

Therefore, ∠AOB = 90°. Hence proved.

Question 10: Prove that the angle between two tangents drawn from an external point is supplementary to the angle subtended at the centre by the segment joining the points of contact.

Proof:

Let PA and PB be two tangents drawn from an external point P to a circle with centre O. Let A and B be their respective points of contact.

The radii drawn to the points of contact are perpendicular to the tangents. Therefore:

∠OAP = 90°

∠OBP = 90°

Now consider quadrilateral OAPB. The sum of its interior angles is 360°.

∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

∠AOB + 90° + ∠APB + 90° = 360°

∠AOB + ∠APB + 180° = 360°

∠AOB + ∠APB = 180°

Therefore, the angle between the tangents, ∠APB, is supplementary to the angle subtended by AB at the centre, ∠AOB. Hence proved.

Question 11: Prove that a parallelogram circumscribing a circle is a rhombus.

Proof:

Let ABCD be a parallelogram whose four sides are tangent to a circle.

For any quadrilateral circumscribing a circle, the sums of the lengths of opposite sides are equal. Therefore:

AB + CD = AD + BC

Since ABCD is a parallelogram:

AB = CD

AD = BC

Substituting these equalities into the tangential quadrilateral relation:

AB + AB = BC + BC

2AB = 2BC

AB = BC

We already know that:

AB = CD and BC = AD

Therefore:

AB = BC = CD = AD

All four sides of the parallelogram are equal. Hence, ABCD is a rhombus. Proved.

Question 12: A triangle ABC circumscribes a circle of radius 4 cm. The point of contact D divides BC into BD = 8 cm and DC = 6 cm. Find AB and AC.

Solution:

Suppose the circle touches AB at E, BC at D and AC at F.

Tangents drawn from the same external point are equal. Therefore:

• BE = BD = 8 cm
• CF = CD = 6 cm
• AE = AF

Let:

AE = AF = x cm

Therefore:

AB = AE + EB = x + 8

AC = AF + FC = x + 6

BC = BD + DC = 8 + 6 = 14 cm

The semiperimeter of triangle ABC is:

s = (AB + BC + AC)/(2)

s = ((x + 8) + 14 + (x + 6))/(2)

s = (2x + 28)/(2)

s = x + 14

The area of a triangle containing an incircle can be calculated as:

Area = radius × semiperimeter

Area = 4(x + 14)

Using Heron's formula:

Area² = s(s - a)(s - b)(s - c)

Here:

• s = x + 14
• a = BC = 14
• b = AC = x + 6
• c = AB = x + 8

Therefore:

s - a = (x + 14) - 14 = x

s - b = (x + 14) - (x + 6) = 8

s - c = (x + 14) - (x + 8) = 6

Using Heron's formula:

Area² = (x + 14)(x)(8)(6)

Area² = 48x(x + 14)

Since Area = 4(x + 14):

[4(x + 14)]² = 48x(x + 14)

16(x + 14)² = 48x(x + 14)

Because x + 14 is positive, divide both sides by 16(x + 14):

x + 14 = 3x

14 = 2x

x = 7 cm

Now calculate the required sides:

AB = x + 8

AB = 7 + 8 = 15 cm

AC = x + 6

AC = 7 + 6 = 13 cm

Therefore, AB = 15 cm and AC = 13 cm.

Question 13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.

Proof:

Let ABCD be a quadrilateral circumscribing a circle with centre O. Suppose the circle touches AB, BC, CD and DA at P, Q, R and S respectively.

Consider triangles OAP and OAS.

• OP = OS, because they are radii of the same circle.
• AP = AS, because tangents drawn from A are equal.
• OA is common to both triangles.

Therefore, triangles OAP and OAS are congruent by the SSS rule.

Hence:

∠POA = ∠AOS

In the same way, we can prove:

∠POB = ∠BOQ

∠QOC = ∠COR

∠ROD = ∠DOS

Let:

• ∠POA = ∠AOS = α
• ∠POB = ∠BOQ = β
• ∠QOC = ∠COR = γ
• ∠ROD = ∠DOS = δ

The angles around point O have a total measure of 360°. Therefore:

2α + 2β + 2γ + 2δ = 360°

2(α + β + γ + δ) = 360°

α + β + γ + δ = 180°

Now:

∠AOB = α + β

∠COD = γ + δ

Therefore:

∠AOB + ∠COD = α + β + γ + δ

∠AOB + ∠COD = 180°

Thus, the angles subtended at the centre by opposite sides AB and CD are supplementary.

Similarly:

∠BOC + ∠DOA = 180°

Therefore, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. Hence proved.