The equation o circle passing through $(3, 4)$ and center at $(1, 2)$ is

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Solution

The equation of circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
Center is $(1, 2)$
The equation of circle is $(x - 1)^{2} + (y - 2)^{2} = r^{2}$ $
It passes through $(3, 4)$
$⟹ (3 - 1)^{2} + (4 - 2)^{2} = r^{2} 2^{2} + 2^{2} = r^{2} r^{2} = 4 + 4 = 8$
So the equation of circle is $(x - 1)^{2} + (y - 2)^{2} = 8 x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 8 x^{2} + y^{2} - 2 x - 4 y - 3 = 0$

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