Find centre and radius for the given equation of the circle.

$x^{2} + y^{2} - 4 x - 8 y - 45 = 0$

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Solution

The given equaion of circle is
$x^{2} + y^{2} - 4 x - 8 y - 45 = 0 ∴ (x^{2} - 4 x) + (y^{2} - 8 y) = 45 \Rightarrow [x^{2} - 4 x + (2)^{2}] + [y^{2} - 8 y + (4)^{2}] = 45 + (2)^{2} + (4)^{2} \Rightarrow (x - 2)^{2} + (y - 4)^{2} = 45 + 4 + 16 \Rightarrow (x - 2)^{2} + (y - 4)^{2} = 65 \Rightarrow (x - 2)^{2} + (y - 4)^{2} = (\sqrt{65})^{2}$
comparing it with $(x - h)^{2} + (y - k)^{=} r^{2},$ we have
h = 2, k = 4 and $r = \sqrt{65}$
Thus co-ordinates of trhe centre is (2,4) and radius is $\sqrt{65}$

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