Find the centre and radius of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$

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Solution

Given: $x^{2} + y^{2} + 2 x + 2 y + 5 = 0$

$\Rightarrow x^{2} + 2 x + 1 - 1 + y^{2} + 2 y + 1 - 1 + 5 = 0$

$\Rightarrow {(x + 1)}^{2} + {(y + 1)}^{2} - 2 - 5 = 0$

$\Rightarrow {(x - (- 1))}^{2} + {(y - (- 1))}^{2} = 8$

which is comparable with ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Here $h = - 1, k = - 1$ and $r = 2 \sqrt{2}$

Hence, the given circle has centre $(- 1, - 1)$ and its radius $= 2 \sqrt{2}$

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

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