If tangents are drawn to the circle $x^2+y^2=12$ at the points where it intersects the circle $x^2+y^2-5 x+3…
- $\left(-6, \frac{18}{5}\right)$
- $\left(6, \frac{18}{5}\right)$
- $\left(-6, \frac{-18}{5}\right)$
- $\left(6, \frac{-18}{5}\right)$
Solution

The equation of the common chord is

Eqs. (i) and (ii) represents the same line. Therefore, $ \begin{aligned} & \frac{h}{5}=\frac{k}{-3}=\frac{-12}{-10} \\ \Rightarrow \quad & h=6, k=\frac{-18}{5} \end{aligned} $ Hence, the required point is $\left(6,-\frac{18}{5}\right)$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)