# How do you find the center and radius of the circle x^2 + y^2 - 4x - 14y + 29 = 0?

Jul 15, 2016

The center will be at $\left(2 , 7\right)$ and the radius is $\sqrt{24}$.

#### Explanation:

This is an intriguing problem that requires several applications of math knowledge. The first of which is just determining what we need to know and what that might look like.

A circle has the generalized equation:

${\left(x + a\right)}^{2} + {\left(y + b\right)}^{2} = {r}^{2}$

Where $a$ and $b$ are the inverses of the circle's center coordinates. $r$, of course, is the radius. So our goal will be taking the equation we're given, and making it have that form.

Looking at the given equation, it seems like our best bet is going to be factoring the two polynomials presented (the one made up of the $x$s and the one made up of the $y$s). It's obvious just from looking at the coefficients of the first degree variables how this will turn out:

${x}^{2} - 4 x \to {\left(x - 2\right)}^{2}$
${y}^{2} - 14 y \to {\left(y - 7\right)}^{2}$

Since those are the only square terms that would give us the appropriate first degree coefficient. But there's a problem!
${\left(x - 2\right)}^{2} = {x}^{2} - 4 x + 4$
${\left(y - 7\right)}^{2} = {y}^{2} - 14 y + 49$

But all we have is the $29$ in the equation. Clearly these constants have been added together to form a single number that doesn't reflect the real radius. We can solve for the real number, $c$, like so:
$4 + 49 + c = 29$
$53 + c = 29$
$c = - 24$

So putting it together we get:
${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} - 24 = 0$
which really is just:
${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = 24$

Now that we have a standard form circle, we can see that the center will be at $\left(2 , 7\right)$ and the radius is $\sqrt{24}$.