Question

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

Answer 1

The equation of the given circle is $x^2+y^2-4x-8y-45=0$.

$(x^2-4x)+(y^2-8y)=45$

$(x-2{)}^2+(y-4{)}^2-4-16=45$

$(x-2{)}^2+(y-4{)}^2=65$

$(x-2{)}^2+(y-4{)}^2=(\sqrt{65}{)}^2$

Thus, the centre of the given circle is $(2,4)$ and radius is $\sqrt{65}$.