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 

Normal at a point to the circle intersect the circle again at the other end of the diameter passing along the normal, 

Centre of the circle $=C\left(3, -4\right).$ Given point $A=\left(8, 8\right)$

Let B be the point at with the normal meets the circle again.

Then midpoint of $AB=C\Rightarrow B=2C-A=2\left(3, -4\right)-\left(8, 8\right)=\left(-2, -16\right)$.