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Tangents are drawn from (-2,0) to y^2=8x...

Tangents are drawn from (-2,0) to `y^2=8x`, then the radius of the circle that would touch these tangents and the corresponding chord of contact can be

A

`4(sqrt(2)+1)`

B

`4(sqrt(2)-1)`

C

`8sqrt(2)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A, B
Point 'P' lies on the dirextrix of `y^(2) = 8x`, slope of PA and PB are 1 and -1 respectively.
Equation of `PA: y = x +2`, Equation of `PB: y =- x -2`, Equation of `AB: x =2`
Let (h,0) be the centre and radius be `'r' rArr (|h+2|)/(sqrt(2)) =(|h-2|)/(1) =r`
`rArr h^(2) - 12h +4 =0 rArr h = 6 +- 4sqrt(2)`
`rArr r = |h -2| = 4 (sqrt(2)-1), 4(sqrt(2)+1)`
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