MHTCET 5 May 2025 Shift 2 Paper

Concept:A diameter of one circle is a chord of another circle, so the radius of the second circle is found using the perpendicular distance from its centre to that chord.Explanation:Consider the given circle: $x^{2} + y^{2} - 4x + 6y - 12 = 0$.Its centre is $C_1(2, -3)$.Its radius is given by:$$r_1 = \sqrt{2^{2} + (-3)^{2} + 12} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$$Hence, the length of its diameter is $2 \times 5 = 10$.This diameter is a chord of circle $S$, whose centre is $C_2(-3, 2)$.The distance between the two centres is:$$C_1 C_2 = \sqrt{(-3 - 2)^{2} + (2 + 3)^{2}} = \sqrt{25 + 25} = \sqrt{50}$$The perpendicular from the centre of a circle to a chord bisects the chord.So half the chord length is $5$.Therefore, the radius of circle $S$ is:$$R_S = \sqrt{(C_1 C_2)^{2} + 5^{2}} = \sqrt{50 + 25} = \sqrt{75} = 5\sqrt{3}$$Answer:The radius of circle $S$ is $5\sqrt{3}$ units, which is option C.