Find the center and radius of a circle from its equation

Key takeaway: For circle equations, the easiest path is to move everything into standard form and then complete the square if needed. Here, the terms are already set up so the center and radius can be read off cleanly.

Start with the given equation:

$x^2+y^2=8x-6y+39$

Move the $x$ and $y$ terms to the left:

$x^2-8x+y^2+6y=39$

Now complete the square for each variable.

For $x^2-8x$:

$\left(\frac{-8}{2}\right)^2=16$

For $y^2+6y$:

$\left(\frac{6}{2}\right)^2=9$

Add both numbers to both sides:

$x^2-8x+16+y^2+6y+9=39+16+9$

$(x-4)^2+(y+3)^2=64$

This matches the standard circle form

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

So the center is

$\boxed{(4,-3)}$

and the radius is

$\boxed{8}$

---

Pitfalls the pros know 👇 A common slip is reading the center as $( -4, 3 )$ because of the signs inside the parentheses. Remember: $(x-h)^2$ means the center’s $x$-coordinate is $h$, not the opposite sign you first see. So $(x-4)^2$ gives $x=4$, and $(y+3)^2$ gives $y=-3$.

What if the problem changes? If the right-hand side had been smaller, the radius would change after completing the square, but the center would stay the same as long as the $x$ and $y$ coefficients stayed the same. If one squared term were missing, then the graph would no longer be a circle and you’d need to interpret it differently.

Tags: standard form, completing the square, radius