# What is the standard form of the equation of a circle with center at (-3, 1) and through the point (2, 13)?

Feb 13, 2016

${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = {13}^{2}$
(see below for discussion of alternate "standard form")

#### Explanation:

The "standard form of an equation for a circle" is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
for a circle with center $\left(a , b\right)$ and radius $r$

Since we are given the center, we only need to compute the radius (using the Pythagorean Theorem)
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(- 3 - 2\right)}^{2} + {\left(1 - 13\right)}^{2}} = \sqrt{{5}^{2} + {12}^{2}} = 13$
So the equation of the circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 1\right)}^{2} = {13}^{2}$

Sometimes what is being asked for is the "standard form of the polynomial" and this is somewhat different.
The "standard form of the polynomial" is expressed as a sum of terms arranged with decreasing degrees set equal to zero.
If this is what your teacher is looking for you will need to expand and rearrange the terms:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x + 9 + {y}^{2} - 2 y + 1 = 169$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} + 6 x - 2 y - 159 = 0$