Find the equation of a circle whose centre is (2, - 1 ) an 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 A $x^{2} + y^{2} - 4 x + 2 y - 4 = 0$
Center $= (2, - 1)$ and $r = 3$ [ Given ]
We know that, center $= (- g, - f)$
So, by comparing we get, $g = - 2$ and $f = 1$
$r = \sqrt{g^{2} + f^{2} - c}$
$\Rightarrow$ $3 = \sqrt{(- 2)^{2} + (1)^{2} - c}$
$\Rightarrow$ $9 = 4 + 1 - c$
$\Rightarrow$ $9 = 5 - c$
$∴$ $c = - 4$
The equation of circle is
$\Rightarrow$ $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$\Rightarrow$ $x^{2} + y^{2} + 2 (- 2) x + 2 (1) y - 4 = 0$ [ Substituting values of $g, f$ and $c$ ]
$\Rightarrow$ $x^{2} + y^{2} - 4 x + 2 y - 4 = 0$

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