Question

What is the Center and radius of x2+67+y2=8y+20x

Answer 1

`x^2+67+y^2=8y+20x`

Let's rewrite the expression as:

`\begin{gathered} x^2+67+y^2-8y-20x=0 \\ so\colon \\ (x-10)^2+(y-4)^2-49=0 \\ (x-10)^2+(y-4)^2=49 \end{gathered}`

Which is the standard equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

Where (h,k) is the coordinates of center of the circle and r is the radius

Therefore, the center is:

`\begin{gathered} (h,k)=(10,4) \\ \end{gathered}`

And the radius is:

`r=\sqrt[]{49}=7`