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

One end of the diameter of the circle $x^2+y^2-6 x-5 y-1=0$ is $(-1,3)$, then the equation of the tangent at the other end of the diameter is

The equation of the circle, concentric with the circle $2 x^2+2 y^2-6 x+8 y+1=0$ and double of its area is