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A circle of non-zero radius has origin as its centre. If it passes through the point of intersection of two curves \(y^2 = 4ax\) and \(x^2 = 4ay\), then its equation is

Question and options

A circle of non-zero radius has origin as its centre. If it passes through the point of intersection of two curves y2=4axy^2 = 4ax and x2=4ayx^2 = 4ay, then its equation is
  1. x2+y2=16a2x^2 + y^2 = 16a^2
  2. x2+y2=4a2x^2 + y^2 = 4a^2
  3. x2+y2=32a2x^2 + y^2 = 32a^2
  4. x2+y2=a2x^2 + y^2 = a^2

Answer and explanation

Correct answer: C - \(x^2 + y^2 = 32a^2\)
y^2 = 4ax and x^2 = 4ay meet at (0,0) and (4a,4a). Non-zero radius means it passes through (4a,4a): r^2 = 16a^2 + 16a^2 = 32a^2.

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  • IPMAT Indore

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