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

Find the non-origin intersection of $y^2 = 4ax$ and $x^2 = 4ay$.

From the second curve, $y = \dfrac{x^2}{4a}$. Substitute into the first:

$\left(\dfrac{x^2}{4a}\right)^2 = 4ax$

$\dfrac{x^4}{16a^2} = 4ax$

$x^4 = 64 a^3 x$

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$x(x^3 - 64a^3) = 0$, so $x = 0$ or $x = 4a$.

For $x = 0$, the point is the origin (giving zero-radius circle, excluded).

For $x = 4a$, $y = \dfrac{(4a)^2}{4a} = 4a$. So the intersection is $(4a, 4a)$.

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The circle centred at origin passing through $(4a, 4a)$ has radius:

$r^2 = (4a)^2 + (4a)^2 = 32 a^2$

Equation: $x^2 + y^2 = 32a^2$

Answer $= x^2 + y^2 = 32a^2$