JEE Main · 2016 · 03 Apr 2016 · Offline · Circle

If one of the diameters of the circle, given by the equation,

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

, is a chord of a circle S, whose centre is at

(- 3, 2)

, then the radius of S is:

(A) $53$ ✓ Correct
(B)5
(C)10
(D) $52$

Answer: (A)

Solution

Completing the square,

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

becomes

(x - 2)^{2} + (y + 3)^{2} = 25

, a circle of centre

(2, - 3)

and radius 5. A diameter of this circle is a chord of S, and that diameter passes through

(2, - 3)

, so the perpendicular distance from S's centre

(- 3, 2)

to the chord is just the distance between the two centres,

(2 + 3)^{2} + (- 3 - 2)^{2} = 50 = 52

. Half the chord equals the first radius, 5, so by

r^{2} = d^{2} + (half-chord)^{2}

r^{2} = 50 + 25 = 75

and

r = 53

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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