Question

If one of the diameters of the circle given by the equation $x^{2}+y^{2}+4 x-6 y-12=0$ is a chord of a circle $S$ whose centre is at $(3,-2)$, find radius of circle $S$. (Take $\sqrt{3}=1.73$ )

Answer 1

Given equation
$\Leftrightarrow(x+2)^{2}+(y-3)^{2}=25$
$\Rightarrow$ Radius $ C_{1} P=5, $
Centre $ C_{1}=(-2,3)$
$\Rightarrow\left|C_{1} C_{2}\right|=\sqrt{(3+2)^{2}+(-2-3)^{2}} $
$=\sqrt{50}$
$C_{2} P =\sqrt{\left(C_{1} C_{2}\right)^{2}+\left(C_{1} P\right)^{2}} $
$=\sqrt{50+25} $
$=5 \sqrt{3} $
$=8.654$
$\Rightarrow$ The required radius $=8.65$