Question

If one of the diameters of the circle given by the equation $x^2+y^2+4x-6y-12=0$ 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

Answer: 8.65

Given equation
$\iff (x+2{)}^2+(y-3{)}^2=25$
$\Rightarrow$ 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 $=8.65$