If the tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+\mathrm{y}^2+16 \mathrm{x}+12 \mathrm{y}+\mathrm{C}=0$, then $\mathrm{C}=$

Finding the tangent equation: For the curve $x^2 = y - 6$, differentiate to find the slope: $2x = \frac{dy}{dx}$, so at $(1,7)$, $m = 2(1) = 2$.

The tangent line is $y - 7 = 2(x - 1)$, which simplifies to $2x - y + 5 = 0$.

Circle analysis: The circle $x^2 + y^2 + 16x + 12y + C = 0$ has center $(-8, -6)$ and radius $R = \sqrt{100 - C}$.

Tangency condition: The distance from the center $(-8, -6)$ to the line $2x - y + 5 = 0$ is $d = \frac{|2(-8) - 1(-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.

Set $d = R$, so $\sqrt{5} = \sqrt{100 - C}$. Squaring both sides yields $5 = 100 - C$, hence $C = 95$.

The value of $C$ is $\boxed{95}$.