Tangent to the parabola $y=x^2+6$ at (1, 7) touches the circle $x^2+y^2+16x+12y+c=0$ at the point

Tangent to parabola $y=x^2+6$ $\Rightarrow$ $\frac{1}{2}\left(2y\right)=x^2+6\text{ at }(1,7)$ is $\frac{1}{2}(y+7)=x\cdot 1+6$

$\Rightarrow$ 2x- y + 5 = 0         ... (1)

Now (1) touches the circle

$x^2+y^2+16x+12y+c=0$... (2)

Therefore (1) and (2) will intersect in two coincide points:

$\therefore x^2+(2x+5{)}^2+16x+12(2x+5)+c=0$ $\Rightarrow$ $5x^2+60x+85+c=0$. Its roots are equal.

$\therefore \alpha +\beta =-\frac{60}{5}=-12$ $\Rightarrow$ $\alpha +\alpha =-12$

$\therefore \alpha =-6$$\Rightarrow$$x=-6$ and  hence from (1) y = - 7

$\therefore$Point of contact is (- 6, - 7).