Find the equation of a circle whose center is (2,-1) and radius is 3

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

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

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

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

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Solution

The correct option is C $x^{2} + y^{2} - 4 x + 2 y - 4 = 0$
Equation of circle with center $(h, k)$ and radius $r$ is written as
$(x - h)^{2} + (y - k)^{2} = r^{2}$
then,
Equation of circle with center $(2, - 1)$ and radius $3$ is written as
$\Rightarrow (x - 2)^{2} + (y + 1)^{2} = 3^{2}$
$\Rightarrow x^{2} + 4 - 4 x + y^{2} + 1 + 2 y = 9$
$∴ x^{2} + y^{2} - 4 x + 2 y - 4 = 0$

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x^{2} + y^{2} - 2 x - 4 y - 20 = 0, x^{2} + y^{2} + 4 x - 2 y + 4

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

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

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

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

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