The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at$\left(8,8\right)$ meets the circle again at the point

$$\begin{gathered} {\mathrm{y}}_1(\mathrm{x}+\mathrm{g})-{\mathrm{x}}_1(\mathrm{y}+\mathrm{f})=0 \\ ({\mathrm{x}}_1, {\mathrm{y}}_1)=(8, 8) \\ (\mathrm{g}, \mathrm{f})=(-3, 4) \\ 8(\mathrm{x}-3)-8(\mathrm{y}+4)=0 \\ \mathrm{x}-3-\mathrm{y}-4=0 \\ \mathrm{x}-\mathrm{y}-7=0 \end{gathered}$$

Verifying options.