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If C1: x^2+y^2=(3+2sqrt(2))^2 is a circl...

If `C_1: x^2+y^2=(3+2sqrt(2))^2` is a circle and `P A` and `P B` are a pair of tangents on `C_1,` where `P` is any point on the director circle of `C_1,` then the radius of the smallest circle which touches `c_1` externally and also the two tangents `P A` and `P B` is
(a) `2sqrt(3)-3` (b) `2sqrt(2)-1` (c) `2sqrt(2)-1` (d) 1

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
Let Q be the centre of the given circle and R be the centre of the smallest circle touching the given circle externally. Then

`AQ = 3 +2 sqrt(2)`
`PQ = 3sqrt(2) +4`
Let radius of smaller circle is r.
`rArr 3 sqrt(2) +4 = QR +RP =3 +2 sqrt(2) +r +r sqrt(2) ( :' /_APQ = (pi)/(4))`
`rArr (sqrt(2)+1) = r (sqrt(2)+1) = r= 1`
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