Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

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 $\sqrt{65}$.