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

JEE MainJEE Main 2018Conic Sections

Solution:

Equation of tangent at $(1,7)$ to curve $x^{2}=y-6$ is
$x-1=\frac{1}{2}(y+7)-6$
$2 x-y+5=0\,\,\,\,\,\,\,...(i)$
Centre of circle $=(-8,-6)$
Radius of circle $=\sqrt{64+36-c}=\sqrt{100-c}$
$\because$ Line (i) touches the circle
$\therefore \left|\frac{2(-8)-(-6)+5}{\sqrt{4+1}}\right|=\sqrt{100-c} $
$ \sqrt{5}=\sqrt{100-c} $
$\Rightarrow c =95 $