## Solution :

Let the equation of circle be \((x-3)^2 + (y-0)^2 + \lambda y\) = 0

As it passes through (1, -2)

\(\therefore\) \((1-3)^2 + (-2)^2 + \lambda (-2)\) = 0

\(\implies\) 4 + 4 – 2\(\lambda\) = 0

\(\implies\) \(\lambda\) = 4

\(\therefore\) Equation of circle is \((x-3)^2 + y^2 + 4y\) = 0

By hit and trial method, we see that point (5, -2) satisfies the equation of circle.

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