<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] 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 $(-6,-7)$.
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