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 `2sqrt(3)-3` (b) `2sqrt(2)-1` `2sqrt(2)-1` (d) 1

Options

A. `2 sqrt(3)-3`
B. `2sqrt(2)-1`
C. `2sqrt(2)-1`
D. 1

Answer

4

Explanation

`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`