# The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point

## 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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