the tangent to curv y= x^2 + 6 at a point 1,7 touches the circle x^2+y^2+16x+12y+c=0 at a point q then the coordinates of q are :

ans = -6,-7

The tangent to the curve $y=x^2+6$ at point (1,7) is given by:
$$\begin{gathered} x*1+6=\frac{1}{2}.(y+7) \\ 2x+12=y+7 \\ 2x-y+5=0 ............\left(1\right) \\ \Rightarrow y=2x+5 ...........\left(2\right) \end{gathered}$$
if equation (1) is the tangent to the circle $x^2+y^2+16x+12y+c=0$
substituting the value of y from eq(2), 
$$\begin{gathered} x^2+(2x+5{)}^2+16x+12*(2x+5)+c=0 \\ {x}^2+4x^2+25+20x+16x+24x+60+c=0 \\ 5x^2+60x+85+c=0 ..........(3) \end{gathered}$$
since the line (1) touches the given circle, then discriminant of eq(3) must be zero:
$$\begin{gathered} {60}^2-4*5*(85+c)=0 \\ 3600-20*(85+c)=0 \\ 180-85-c=0 \\ c=180-85=95 \end{gathered}$$
thus equation (3) can be rewritten as $5x^2+60x+85+95=0$
$$\begin{gathered} \Rightarrow 5x^2+60x+180=0 \\ {x}^2+12x+36=0 \\ \Rightarrow (x+6{)}^2=0 \\ \Rightarrow x=-6 \end{gathered}$$
for $x=-6 , we have y=2*(-6)+5$
$$\begin{gathered} y=-12+5 \\ y=-7 \end{gathered}$$
if the point of contact is Q, then the coordinates of Q are $(-6,-7)$

hope this helps you