# What is the standard form of the equation of a circle with center (1,2) intersects x-axis at -1 and 3?

Jan 5, 2016

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 8$

#### Explanation:

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

In the case the radius is the distance between the center $\left(1 , 2\right)$ and one of the points on the circle; in this case we could use either of the x-intercepts: $\left(- 1 , 0\right)$ or $\left(3 , 0\right)$
to get (using $\left(- 1 , 0\right)$):
$\textcolor{w h i t e}{\text{XXXXXXXX}} r = \sqrt{{\left(1 - \left(- 1\right)\right)}^{2} + {\left(2 - 0\right)}^{2}} = 2 \sqrt{2}$

Using $\left(a , b\right) = \left(1 , 2\right)$ and ${r}^{2} = {\left(2 \sqrt{2}\right)}^{2} = 8$
with the general standard form gives the answer above.