Q49 of 63 Page 202

Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is

As the centre lies on y – axis, then the centre is (0, k)


radius = ‘k’ units


Using Distance Formula, {Between (2, 3) & (0, k)}






Equating the equations,
r = k &


k =


k2 = k2 - 6k + 13


6k = 13



Centre =


Since, the equation of a circle having centre (h,k), having radius as "r" units, is


(x – h)2 + (y – k)2 = r2





3x2 + 3y2 - 13y = 0

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