Find the centre & radius of the circle whose equation is $x^{2} + y^{2} - 4 x + 6 y + 12 = 0$

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Solution

Given $x^{2} + y^{2} - 4 x + 6 y + 12 = 0$

Comparing it with the general equation $x^{2} + y^{2} - 2 g x + 2 f y + c = 0$ we have

$2 g = - 4$ $\Rightarrow$ $g = - 2$

$2 f = 6$ $\Rightarrow$ $f = 3$

& $c = 12$

$∴$ centre is (-g, -f) i.e. (2, -3)

and radius = $\sqrt{g^{2} + f^{2} - c} = \sqrt{{(- 2)}^{2} + {(3)}^{2} - 12} = 1$

Centre $(2, - 3)$ and radius= $1$

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