# How do find the standard equation of the circle with center (3,5) and passes through (-1,-2)?

May 4, 2017

The standard Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and r is the radius.

#### Explanation:

Substitute the center $\left(3 , 5\right)$ into equation [1]:

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {r}^{2} \text{ [2]}$

Substitute the point $\left(- 1 , - 2\right)$ into equation [2] and solve for r:

${\left(- 1 - 3\right)}^{2} + {\left(- 2 - 5\right)}^{2} = {r}^{2}$

${\left(- 4\right)}^{2} + {\left(- 7\right)}^{2} = {r}^{2}$

$16 + 49 = {r}^{2}$

${r}^{2} = 65$

$r = \sqrt{65}$

Substitute $\sqrt{65}$ for r into equation [2]:

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(\sqrt{65}\right)}^{2} \text{ [3]}$

Note: Please understand that I write the right side as ${\left(\sqrt{65}\right)}^{2}$, instead of 65, because anyone asked for the radius will not make the mistake of reporting 65.