Answer:
Option D
Explanation:
Key Idea Equation of tangent to the curve
x2=4ay at ( x1,y1) is $xx_{1}=4a( \frac{y+y_{1}}{2})$
Tagent to the curve $x^{2}=y-6$ at ( 1,7)
$x=\frac{y+7}{2}-6$
$\Rightarrow$ 2x-y+5=0 .........( i)
Equation of circle is
$x^{2}+y^{2}+16x+12y+c=0$
Centre ( -8,-6)
$r=\sqrt{8^{2}+6^{2}-c}=\sqrt{100-c}$
Since , line 2x-y+5=0 also touches the circle.
$\therefore$ $\sqrt{100-c}= \mid \frac{2( -8)-( -6)+5}{\sqrt{2^{2}+1^{2}}}\mid$
$\therefore$ $\sqrt{100-c}= \mid \frac{-16+6+5}{\sqrt{5^{}}}\mid$
$\therefore$ $\sqrt{100-c}= \mid -\sqrt{5}\mid$
$\therefore$ 100-c=5 $\therefore$ c=95
