Question

If one 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 $(2,–3),$ the radius of $S$ is

• A. $\sqrt{41}\text{unit}$
• B. $3\sqrt{5}\text{unit}$
• C. $5\sqrt{2}\text{unit}$
• D. $2\sqrt{5}\text{unit}$

Answer 1

The correct option is A $\sqrt{41}\text{unit}$

$x^2+y^2+4x+6y–12=0\Rightarrow (x+2{)}^2+(y+3{)}^2=25$


$C(-2,-3),r=CP=5$
$CG=\sqrt{16+0}=4G{P}^2=C{G}^2+C{P}^2\Rightarrow G{P}^2=16+25\Rightarrow GP=\sqrt{41}$