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:
A.
$$5$$
B.
$$10$$
C.
$$5\sqrt 2 $$
D.
$$5\sqrt 3 $$
Answer :
$$5\sqrt 3 $$
Solution :
Centre of $$S:O\left( { - 3,\,2} \right)$$ centre of given circle $$A\left( {2,\, - 3} \right)$$
$$ \Rightarrow OA = 5\sqrt 2 $$
Also $$AB=5$$ ($$\because AB = r$$ of the given circle)
$$ \Rightarrow $$ Using Pythagoras theorem in $$\Delta OAB$$
$$r = 5\sqrt 3 $$
Centre of $$S:O\left( { - 3,\,2} \right)$$ centre of given circle $$A\left( {2,\, - 3} \right)$$
$$ \Rightarrow OA = 5\sqrt 2 $$
Also $$AB=5$$ ($$\because AB = r$$ of the given circle)
$$ \Rightarrow $$ Using Pythagoras theorem in $$\Delta OAB$$
$$r = 5\sqrt 3 $$