# How do you write an equation of a circle with Center (-1, 15) Point on Circle (2, 13)?

Nov 24, 2016

${\left(x + 1\right)}^{2} + {\left(y - 15\right)}^{2} = 13$

#### Explanation:

If 'r' is the radius of the circle then its equation would be

${\left(x + 1\right)}^{2} + {\left(y - 15\right)}^{2} = {r}^{2}$

Since point (2,13) lies on the circle, it must satisfy the above equation. Thus

${\left(2 + 1\right)}^{2} + {\left(13 - 15\right)}^{2} = {r}^{2}$
$9 + 4 = 13 = {r}^{2}$

Thus the equation of the circle becomes
${\left(x + 1\right)}^{2} + {\left(y - 15\right)}^{2} = 13$

Nov 24, 2016

#### Explanation:

Use the standard equation of a circle:

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

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

You are not given r, therefore, you must substitute the given point and the center into the standard form, and then compute value of r.

${\left(2 - - 1\right)}^{2} + {\left(13 - 15\right)}^{2} = {r}^{2}$

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

${r}^{2} = 9 + 4$

${r}^{2} = 13$

$r = \sqrt{13}$

The equation of the circle is:

${\left(x - - 1\right)}^{2} + {\left(y - 15\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$