# How do you write an equation of a circle with center (2, 4) and passes through the point (6,7)?

Jul 25, 2016

${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 25$

#### Explanation:

If the circle has its center at $\left({x}_{c} , {y}_{c}\right) = \left(2 , 4\right)$ and passes through $\left(6 , 7\right)$
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(6 - 2\right)}^{2} + {\left(7 - 3\right)}^{2}} = 5$
Since the general equation for a circle with center $\left({x}_{c} , {y}_{c}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = {5}^{2}$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 4 x - 8 y - 5 = 0$