If the tangent at (1, 7) to the curve, x2=y−6 touches the circle x2+y2+16x+12y+c=0 then the value of c is

If the tangent at (1, 7) to the curve, $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ then the value of $c$ is

A
95
B
195
C
185
D
85

Solution:

The equation of tangent at (1, 7) to $x^{2} = y - 6$ is $2 x - y + 5 = 0$

The perpendicular distance of Centre (-8, -6) to the line $2 x - y + 5 = 0$ should be equal to the radius of the circle.

$∴ \sqrt{64 + 36 - c} = \frac{| - 16 + 6 + 5 |}{\sqrt{5}}$

$\Rightarrow \sqrt{5} = \sqrt{100 - c} \Rightarrow c = 95$

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