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If r1a n dr2 are the radii of the smalle...

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`

A

`31//4`

B

`41//4`

C

`41//3`

D

17

Text Solution

Verified by Experts

The correct Answer is:
2
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)`
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