Question

If one of the diameter 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), then the radius of S is:
(A) $5\sqrt{3}$
(B) 5

Answer 1

We have,

Equation of circle

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

Comparing that,

$x^2+y^2+2gx+2fy+c=0$

$g=-2,f=3,c=-12$

Then,

Radius of circle

$=\sqrt{g^2+f^2-c}$

$=\sqrt{{(-2)}^2+{\left(3\right)}^2-(-12)}$

$=\sqrt{4+9+12}$

$=\sqrt{25}$

$=5$

Hence, this is the answer.