The normal drawn at $P(-1,2)$ on the circle $x^2+y^2-2 x-2 y-3=0$ meets the circle at another point $Q$.…

The normal drawn at $P(-1,2)$ on the circle $x^2+y^2-2 x-2 y-3=0$ meets the circle at another point $Q$. Then, the coordinates of $Q$ are
  1. $(3,0)$
  2. $(-3,0)$
  3. $(2,0)$
  4. $(-2,0)$

Solution

Equation of circle, $x^2+y^2+2 x-2 y-3=0$
Its centre is $(1,1)$ Given that, $\therefore \quad$ PQ being normal to the circle represent the diameter of circle and $\mathrm{O}$ is the mid-point of PQ. $\begin{aligned} & \therefore \quad \frac{x-1}{2}=1, \frac{y+2}{2}=1 \\ & x=3, y=0 \end{aligned}$ So, coordinate of $Q=(3,0)$

Asked in: AP EAMCET 2016

Practice more Circle questions on Aicharya