If one of the diameters of the circle given by the equation x^2+y^2-4x+6y-12=0, is a chord…

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

Accepted Answer

Center of $S$ : $O(-3, 2)$ center of given circle $A(2, -3)$

$$ \Rightarrow OA = 5\sqrt 2 $$

Also $AB=5$ (as $AB=r$ of the given circle)

$$ \Rightarrow $$ Using pythagoras theorem in $$\Delta OAB$$

$$r = 5\sqrt 3 $$

Solutions · 1