The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at... 
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...

What I tried...

The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\frac{dy}{dx}\Bigg|_{(1,7)}=2\tag{Slope of the line tangent to the parabola}$$
So the equation of the line is $2x-y+5=0\implies y=2x+5$
Substituting this in the equation of circle to find the point of intersection of the line with the circle, we get,
$$x^2+(2x+5)^2+16x+12(2x+5)+c=0$$
Solving this, I get a complicated equation and then the answer comes out in terms of $c$ but the actual answer does not contain $c$ at all.
I would prefer a more analytical/geometrical approach if possible
 A: Because the line "touches" the circle, there is only one point of intersection with the circle. Therefore, the equation,
$$
x^2+(2x+5)^2+16x+12(2x+5)+c=0 \implies  5 x^2 + 60 x + 85 + c= 0
$$
has one solution.
So the determinant $\Delta = 3600 - 4\cdot5\cdot(85+c)=0$. Ergo, $c=95$.
And solving the equation $5x^2 + 60x +180=0$ gives the solution $(-6,-7)$.
A: By rearranging the equation of the circle, we quickly find that the centre of the circle is $(-8,-6)$ (as you have illustrated).
Let $O$ be the centre of the circle, and $P$ be the point where the line $y=2x+5$ and circle touch. Since the line and circle touch at $P,$ the line through $O$ and $P$ must be the line perpendicular to $y=2x+5$ at $P$ (that is, the radius is perpendicular to the tangent).
So now we're looking for a line perpendicular to $y=2x+5$ and passing through $O=(-8,-6).$ By vector geometry, this line has parametric form $r(t) = (-8,-6) + t(-2,1),$ i.e., $r(t) = (-2t-8,t-6).$ Now set $x=-2t-8$ and $y=t-6$ in $y=2x+5$ and solve for $t$: we find $t=1,$ i.e., $P=(-6,-7).$
A: 
As you've simplified the question, we need to find
a point, where the line
\begin{align} 
y&=2x+5
\tag{1}\label{1}
\end{align}
touches the circle of unknown radius
centered at $O=(-8,-6)$.
A convenient point on the tangent line below the center $O$ is $A(-8,-11)$, $|OA|=5$.
Let $\phi=\angle T_cOA$,
\begin{align}
\phi&=\arctan2
=\arccos\tfrac{\sqrt5}5
=\arcsin\tfrac{2\sqrt5}5
\tag{2}\label{2}
.
\end{align}
Then the radius of the circle
\begin{align}
r=|OT_c|&=|OA|\cdot\cos\phi
=\sqrt5
\tag{3}\label{3}
,\\
T_c&=O+r\cdot(\sin\phi,\,-\cos\phi)
\\
&=O+(2,-1)=(-6,-7)
.
\end{align}
