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JEE Main 2017 · Mathematics · Straight Lines and Circles · medium

The equation of the circle passing through the points (0,0), (4,0) and (0,6) is

Single correct option

The equation of the circle passing through the points (0,0)(0,0), (4,0)(4,0) and (0,6)(0,6) is
  1. A. x2+y2−4x−6y=0x^2+y^2-4x-6y=0
  2. B. x2+y2+4x+6y=0x^2+y^2+4x+6y=0
  3. C. x2+y2−6x−4y=0x^2+y^2-6x-4y=0
  4. D. x2+y2−4x−6y+12=0x^2+y^2-4x-6y+12=0
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Answer: (A) x2+y2−4x−6y=0x^2+y^2-4x-6y=0

The circle passes through the origin, so write x2+y2+2gx+2fy=0x^2+y^2+2gx+2fy=0. Using (4,0)(4,0): 16+8g=016+8g=0 gives g=−2g=-2. Using (0,6)(0,6): 36+12f=036+12f=0 gives f=−3f=-3. Hence x2+y2−4x−6y=0x^2+y^2-4x-6y=0, with centre (2,3)(2,3) and radius 13\sqrt{13}. It is the circle on the hypotenuse of the right triangle as diameter.
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