Tangents are drawn to the circle x²+y²= 16 at the points where it is met by the circle x²+y²-5x+3y-2=0, the point of intersection of these tangetnts is ?

Dear Student,

The equation of the chord of contact to the circle $C_1: x^2+y^2=16$ at a point $\left(a, b\right)$ is given as $ax+by=16.$
The equation of the two circles are $C_1: x^2+y^2=16$ and $C_2: x^2+y^2-5x+3y-2=0$.
The equation of their common chord is given as $C_2-C_1=0$.
$$\begin{gathered} \Rightarrow x^2+y^2-5x+3y-2-\left(x^2+y^2-16\right)=0 \\ \Rightarrow x^2+y^2-5x+3y-2-x^2-y^2+16=0 \\ \Rightarrow -5x+3y+14=0 \\ \Rightarrow 5x-3y=14 \end{gathered}$$
Since, the tangents drawn to the circle $C_1: x^2+y^2=16$ at the points where it is met by the circle $C_2: x^2+y^2-5x+3y-2=0$ so the chord of contact of the circle $C_1: x^2+y^2=16$ is same as their common chord.
$$\begin{gathered} \Rightarrow ax+by=16 and 5x-3y=14 represent the same line. \\ \Rightarrow \frac{a}{5}=\frac{b}{-3}=\frac{16}{14} \\ \frac{a}{5}=\frac{16}{14}\Rightarrow a=\frac{40}{7} \\ \frac{b}{-3}=\frac{16}{14}\Rightarrow b=\frac{-24}{7} \\ \left(a, b\right)=\left(\frac{40}{7}, \frac{-24}{7}\right) \end{gathered}$$
Thus, the point of intersection of these tangents is $\left(\frac{40}{7}, \frac{-24}{7}\right)$.
Regards,