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JEE Main 2016

JEE Main 2016 Mathematics question on Circles

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 SS, whose centre is at (−3,2)(-3, 2), then the radius of SS is
  1. A.
    525\sqrt{2}
  2. B.
    535\sqrt{3}
  3. C.
    55
  4. D.
    1010
Show answer and solution

Correct option: B

Solution

The given circle has centre (2, -3) and radius √(4+9+12) = 5. A diameter of this circle is a chord of S, and S has centre (-3, 2). The distance from (-3, 2) to (2, -3) is √(25+25) = 5√2. Since the diameter (length 10, so half-length 5) is a chord of S, by the Pythagorean theorem the radius of S = √((5√2)² + 5²) = √(50+25) = √75 = 5√3.

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