Question

Find the center and radius of the circle whose equation is given by $-x^2-y^2+8x=0$

Answer 1

An equation of the circle with center $(h,k)$ and radius $r$ is

$(x-h{)}^2+(y-k{)}^2=r^2$

Consider, $-x^2-y^2+8x=0$, it can be rewritten as:

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

$x^2+y^2-8x+16=0+16$

implies $(x-4{)}^2+(y+0{)}^2=16$

So, comparing $(x-4{)}^2+(y+0{)}^2=16$ with the above equation of circle we get:

$h=4$, $k=0$ and $r=4$

Therefore, center of the circle is $(4,0)$ and radius of the circle is $r=4$.