Question:
Find the centre and radius of the circle $x^{2}+y^{2}-4 x-8 y-45=0$
Solution:
The equation of the given circle is $x^{2}+y^{2}-4 x-8 y-45=0$.
$x^{2}+y^{2}-4 x-8 y-45=0$
$\Rightarrow\left(x^{2}-4 x\right)+\left(y^{2}-8 y\right)=45$
$\Rightarrow\left\{x^{2}-2(x)(2)+2^{2}\right\}+\left\{y^{2}-2(y)(4)+4^{2}\right\}-4-16=45$
$\Rightarrow(x-2)^{2}+(y-4)^{2}=65$
$\Rightarrow(x-2)^{2}+(y-4)^{2}=(\sqrt{65})^{2}$, which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=2, k=4$, and $r=\sqrt{65}$.
Thus, the centre of the given circle is $(2,4)$, while its radius is $\sqrt{65}$.
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