If `r_1a n dr_2` are the radii of the smallest and the largest circles, respectively, which pass though (5, 6) and touch the circle `(x-2)^2+y^2=4,` then `r_1r_2` is (a) `4/(41)` (b) `(41)/4` `5/(41)` (d) `(41)/6`

Options

A. `31//4`
B. `41//4`
C. `41//3`
D. 17

Answer

2

Explanation

The given circle is `(x-2)^(2)+y^(2)=4`, the centre is (2,0), and the radius is 2.
Therefore, distance between `(2,0)` and (5,6) is `sqrt(9+36)=3 sqrt(5)`. Then `r_(1)=(3sqrt(5)-2)/(2)`
and `r_(2)=(d_(2))/(2)=(3sqrt(5)+2)/(2)`
`:. r_(1)r_(2)=(41)/(4)`