Question

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

Answer 1

_Given that _

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

Then we know that The standard equation of the circle is $(x - h)^2 + (y -k)^2 = r^2$

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

$⇒x^2+2^2-4x+y^2+4^2-8y-20-45=0$

$⇒x^2+2^2-4x+y^2+4^2-8y-65=0$

$⇒(x-2)^2+(y-4)^2=(√65)^2$

Now comparing the above equation with the standard equation we get

The circle of the equation is $(2,4)$ and the radius is $√65$

So by comparing the standard equation with the given equation, we get

$h=-5 , k=3$ and $r=6$

Hence the center of the circle is $(-5,3)$, and the radius is $6$.