Equation of the circle center at the point (1, 2) and passing through the point (0, 2) is

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

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x^{2} + y^{2} + 2 x = 0

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{13 x}^{2} + {13 y}^{2} - 13 x - 61 y + 9 = 0

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none of these

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Solution

The correct option is A $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$
The radius is distance between points
$r = \sqrt{(1 - 0)^{2} + (2 - 2)^{2}} = \sqrt{1 + 0} = 1$
Equation of circle is
$(x - 1)^{2} + (y - 2)^{2} = 1 x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 1 x^{2} + y^{2} - 2 x - 4 y + 4 = 0$

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Similar questions

x^{2} + y^{2} + 2 y - 3 = 0,

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines
$x^{2} - y^{2} - 2 x + 4 y - 3 = 0,$ is

x + 2 y - 3 = 0

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

x^{2} + y^{2} = 4

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

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