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 $(2,-3)$, the radius of $S$ is

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

Answer 1

Answer: (A)

Given, equation of circle is
$x^2+y^2+4x+6y-12=0$
Centre $C(-2,-3)$
and radius $=\sqrt{(-2{)}^2+(-3{)}^2+12}=5$

$x^2+y^2+4x+6y-12=0$
$\therefore C_1{C}_2=\sqrt{(2+2{)}^2+(-3+3{)}^2}$
$=\sqrt{(4{)}^2+(0{)}^2}=4$
$\therefore$ Radius of circle, $S$
$=\sqrt{(4{)}^2+(5{)}^5}$
$=\sqrt{16+25}$
$=\sqrt{41}$ units