Let S≡0.S1≡0 be two externally touching circles whose centres are at C1,Crespectively, whereS≡0 is the smaller one. Let the common tangent drawn from a point P, touches S≡0 at A, S1≡0 at B where PA=4, PB=8 and P,C1,C2 are collinear . Let the common tangent to S≡0,S1≡0 at their point of contract Q meets AB at C. Which of the following is/are correct?

since ∆PAC1,∆PBC2 are similarAC1BC2=PAPB⇒r2=2r1PC2−PC1=C1C2=r1 +r2=3r1⇒64+4r12−16+r12=3r1 2 16+r12-16+r12=3r1 ⇒16+r12=9r12⇒r1=2,r2=22CQ=AC=CB=2_ C1C2=32 the length of the tangents drawn from C to the circles are equal ,so CQ=AC=CB,and AB=PB-PA=8-4=4 Area of ∆C1C2C =12C1C2CQ=1232 ×2=32

Let S≡0.S1≡0 be two externally touching circles whose centres are at C1,Crespectively, whereS≡0 is the smaller one. Let the common tangent drawn from a point P, touches S≡0 at A, S1≡0 at B where PA=4, PB=8 and P,C1,C2 are collinear . Let the common tangent to S≡0,S1≡0 at their point of contract Q meets AB at C. Which of the following is/are correct?

since ∆PAC1,∆PBC2 are similarAC1BC2=PAPB⇒r2=2r1PC2−PC1=C1C2=r1 +r2=3r1⇒64+4r12−16+r12=3r1 2 16+r12-16+r12=3r1 ⇒16+r12=9r12⇒r1=2,r2=22CQ=AC=CB=2_ C1C2=32 the length of the tangents drawn from C to the circles are equal ,so CQ=AC=CB,and AB=PB-PA=8-4=4 Area of ∆C1C2C =12C1C2CQ=1232 ×2=32

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$Let S \equiv 0 . S^{1} \equiv 0 be two externally touching circles whose centres are at C_{1}, C$ respectively, where $S \equiv 0$ is the smaller one. Let the common tangent drawn from a point P, touches S $\equiv 0$ at A, $S^{1} \equiv 0$ at B where PA=4, PB=8 and $P, C_{1}, C_{2} are collinear . Let the common tangent to S \equiv 0, S^{1} \equiv 0$ at their point of contract Q meets AB at C. Which of the following is/are correct?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$A C_{1}, C Q, B C_{2} are in A.P$

$Area of triangle C_{1} C C_{2} is 3 \sqrt{2} Sq.units$

$Area of smaller circle is 2 π$

$Area of S^{1} = O is 4 π$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sin c e ∆ P A C_{1}, ∆ P B C_{2} a r e s i m i l a r$

$\begin{aligned} \frac{A C_{1}}{B C_{2}} = \frac{P A}{P B} \\ \Rightarrow r_{2} = 2 r_{1} \end{aligned}$

$P C_{2} - P C_{1} = C_{1} C_{2} = r_{1_{}} + r_{2} = 3 r_{1} \Rightarrow \sqrt{64 + 4 r_{1}^{2}} - \sqrt{16 + r_{1}^{2}} = 3 r_{1} 2 \sqrt{16 + r_{1}^{2}} - \sqrt{16 + r_{1}^{2}} = 3 r_{1}$

$\begin{aligned} \Rightarrow 16 + r_{1}^{2} = 9 r_{1}^{2} \Rightarrow r_{1} = \sqrt{2}, r_{2} = 2 \sqrt{2} \\ CQ = AC = CB = \underset{}{\underline{2}} C_{1} C_{2} = 3 \sqrt{2} \end{aligned}$ $(t h e l e n g t h o f t h e \tan g e n t s d r a w n f r o m C t o t h e c i r c l e s a r e e q u a l, s o C Q = A C = C B, a n d A B = P B - P A = 8 - 4 = 4)$ $A r e a o f ∆ C_{1} C_{2} C_{} = \frac{1}{2} (C_{1} C_{2}) (C Q) = \frac{1}{2} 3 \sqrt{2} \times 2 = 3 \sqrt{2}$