Question

Find the center and radius of the circle $x^2+y^2-4x-8y-45=0$

Answer 1

Given equation of circle is $x^2+y^2-4x-8y-45=0$
$\Rightarrow (x^2-4x)+(y^2-8y)=45$
$\Rightarrow$ $\{x^2-2\left(x\right)\left(2\right)+2^2\}+\{y^2-2\left(y\right)\left(4\right)+4^2\}-4-16=45$
$\Rightarrow {(x-2)}^2+{(y-4)}^2=65$
$\Rightarrow$ ${(x-2)}^2+{(y-4)}^2={\left(\sqrt{65}\right)}^2$
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 $\sqrt{65}$