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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` `2sqrt(2)-1` (d) 1

A

`2 sqrt(3)-3`

B

`2sqrt(2)-1`

C

`2sqrt(2)-1`

D

1

Text Solution

Verified by Experts

The correct Answer is:
4
`AQ=3+2sqrt(2)`
`PQ =3sqrt(2)+4`

Let r be the required radius. Then,
`3sqrt(2)+4=3+2sqrt(2)+r +rsqrt(2)`
`sqrt(2)+1=r(1+sqrt(2))` or `r=1`
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