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
$$5\sqrt 3 $$
Explanation
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 $$\Delta OAB$$
$$r = 5\sqrt 3 $$
2
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
The centres of those circles which touch the circle, $${x^2} + {y^2} - 8x - 8y - 4 = 0$$, externally and also touch the $$x$$-axis, lie on: