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Q.

Suppose $S_{1}$ and $S_{2}$ are two unequal circles; AB and CD are the direct common tangents to the circles. A transverse common tangent PQ cuts AB in R and CD in S. If AB=10, then RS is

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a

8

b

9

c

10

d

11

answer is C.

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Detailed Solution

We know the length of two tange form a point to circle are of equal length
⇒RQ = RS and SP = SC = y
Let RQ = RS = y
SP = SC = y
Given AB = 10
AB = AR + RB
AR = AB – RB
RP = 10 – r
PR = PQ – RQ
PQ = PR – RQ
10 – x – x
= 10 – 27
SQ = SP +PQ = y+10-2x = SD
AB = CD = 10
CD = CS + SD
CD – CS = SD
SD= 10 -2x +y
CD = CS + SD
10 = y + 10 – 2x + y
10 = 10 – 2x +2y

x = y
RS = RP + PQ + PS
PQ = x PQ = 10 – x PS = y
Rs = x +10 – 2x + y
= 10 - x + y
X = y
RS = 10 – x + y
RS = 10

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