$$\begin{gathered} Given ellipse \frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1 \\ where a^2=16,b^2=9 \\ and centre\left(C\right)=(4,3) \end{gathered}$$

Image of C(4, 3) under x – y – 2 = 0 is (5, 2)
Required ellipse is S$\equiv$ $\frac{(\mathrm{x}-5{)}^2}{9}+\frac{(\mathrm{y}-2{)}^2}{16}=1$

locus of point of intersection of perpendicular tangents is  nothing but  director circle
$$So,it's equation is $x^2+y^2=9+16=25$