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
$x^{2}+y^{2}+4 x+6 y-12=0$
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