<s>[INST] 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
 [/INST] 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., $P=(-6,-7).$
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