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Let the points $P(a,b)$ and $Q(2, 1)$ lie on a parabola $x^2 = 4y$. If a circle is drawn with $PQ$ as diameter touching the parabola $x^2 = 4y$ at $R$. then we have to prove that the value of $a^2 - 4a$ is $60$ .

I tried using $a^2 = 4b$.

Equation of circle is $(x-2)(x-a) + (y-1)(y-b)=0$.

But I don't know how to proceed.

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Since the parabola and circle are touching each other at R we could obtain another equation by equating the equation of tangent (write T = 0 for both curves) at that point. The line PR and QR also intersect at right angles if that helps.

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