The tangent to the curve $y=x^2+6$ at a point P(l, 7) touches the circle $x^2+y^2+16x+12y+c=0$ at a point 

Q. Then the co-ordinates of Q are 

Curve, y $=m^2+6$

${\Rightarrow }^2=-6$......(1)

Vertex: $\left(0,6\right),a=\frac{1}{4}$

T at P $(1,7)\Rightarrow x=\frac{y+7}{2}-6$

$2x=y-5$......(2)          $(\mathrm{m}=2)\left({\mathrm{c}}^{\mathrm{\prime }}=5\right)$

Circle: $x^2+y^2+16x+12y+c=0$......(3) $\left(\text{ radius }={\sqrt{8}}^2+6^2-c\right)$

If (2) is tangent to (3),

$\Rightarrow \left(\text{ radius }\right)\left(\sqrt{1}+{\mathrm{m}}^2\right)={\mathrm{c}}^{\mathrm{\prime }}$

$\Rightarrow (5{)}^2=\left[1+(2{)}^2\right][100-c]$

$\Rightarrow 5=100-c$

$\Rightarrow >\mathrm{c}=95$

$\Rightarrow a=\sqrt{5}$

PO contact: $\left(\pm \frac{am}{\sqrt{1+m^2}},\pm \frac{a}{\sqrt{1+m^2}}\right)$

$\Rightarrow (-6,-7)$

Hence, option A is correct.