Explanation

(d)
The tangent at (1, 7) to the parabola $x^{2} = y - 6$ is
$x(1) = \frac{1}{2}(y + 7) - 6$
[replacing $x^{2} \rightarrow xx_{1}$ and $2y \rightarrow y + y_{1}\rbrack$
$\Rightarrow \ \ 2x = y + 7 - 12$
$\Rightarrow \ \ y = 2x + 5$ …(i)
Which is also tangent to the circle
$x^{2} + y^{2} + 16x + 12y + c = 0$
$ie,\ x^{2} + (2x + 5)^{2} + 16x + 12(2x + 5) + c = 0$
or $5x^{2} + 60x + 85 + c = 0$
Must have equal roots
$\Rightarrow \ \alpha = \beta$ for above equation $ie$,
$\Rightarrow \ \alpha + \beta = - \frac{60}{5}$
or $\alpha = - 6\ \ \ \ \ (as\ \alpha = \beta)$
$\therefore\ x = - 6\$
and $y = 2x + 5 = - 7$
$\Rightarrow$ Point of contact is$\ ( - 6, - 7)$