Find the coordinates of the centres and the radii of the circles whose equation is:
$x^{2} + y^{2} - 4 x - 8 y = 41$

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Solution

Given equation of circle is, $x^{2} + y^{2} - 4 x - 8 y = 41$
$\Rightarrow (x^{2} - 4 x) + (y^{2} - 8 y) = 41$
$\Rightarrow (x^{2} - 4 x + 4) + (y^{2} - 8 y + 16) = 41 + 4 + 16$
$\Rightarrow (x - 2)^{2} + (y - 4)^{2} = (\sqrt{61})^{2}$
Comparing with $(x - h)^{2} + (y - k)^{2} = r^{2}$
We get center of the circle as $(2, 4)$ , and radius $\sqrt{61}$

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