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MathematicsSingle correct+4 / -1Question 80 of 90

JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 80

If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0x^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 :

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Answer and worked solution

Official answer: B

The given circle has centre O1(2,−3)O_1(2,-3) and radius 5, so its diameter is a chord of S of length 10 whose midpoint is O1O_1. The line from S's centre C(−3, 2) to the midpoint of a chord is perpendicular to it, so
r2=CO12+52=(25+25)+25=75,r=53.r^2=CO_1^2+5^2=(25+25)+25=75,\qquad r=5\sqrt3.

Topics:Equation of a circle
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