# How do you find the equation of the circle with center at (3, 2) and through the point (5, 4)?

Jun 18, 2016

The equation is ${x}^{2} + {y}^{2} - 6 x - 4 y + 5 = 0$.

#### Explanation:

The generic equation of a circle with center $\left({c}_{x} , {c}_{y}\right)$ and radius $r$ is:

${\left(x - {c}_{x}\right)}^{2} + \left(y - {c}_{y}\right) = {r}^{2}$

in our case

${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = {r}^{2}$

What is missing here is the radius, but we can find it substituting the point $\left(5 , 4\right)$

${\left(5 - 3\right)}^{2} + {\left(4 - 2\right)}^{2} = {r}^{2}$

${2}^{2} + {2}^{2} = {r}^{2}$

$4 + 4 = {r}^{2}$

$r = \sqrt{8}$

The final equation is

${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 8$ or, expanding the squares,

${x}^{2} + {y}^{2} - 6 x - 4 y + 5 = 0$.