# How do you write an equation for a circle given center (8,-9) and passes through (21,22)?

Dec 12, 2016

#### Explanation:

The standard 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 point, and r is the radius.

We are given that the center point is $\left(8 , - 9\right)$, therefore, we substitute 8 for h and -9 for k into equation [1]:

${\left(x - 8\right)}^{2} + {\left(y - - 9\right)}^{2} = {r}^{2} \text{ [2]}$

NOTE: One can write ${\left(y - - 9\right)}^{2}$ as ${\left(y + 9\right)}^{2}$ but doing so can cause errors when attempting to obtain the center point. Therefore, I do not recommend it.

To find the value of r, substitute the given point, 21 for x and 22 for y, into equation [2]:

${\left(21 - 8\right)}^{2} + {\left(22 - - 9\right)}^{2} = {r}^{2}$

${\left(13\right)}^{2} + {\left(31\right)}^{2} = {r}^{2}$

$169 + 961 = {r}^{2}$

${r}^{2} = 1130$

$r = \sqrt{1130}$

Substitute $\sqrt{1130}$ for r in equation [2]:

${\left(x - 8\right)}^{2} + {\left(y - - 9\right)}^{2} = {\left(\sqrt{1130}\right)}^{2} \text{ [3]}$

Equation [3] is the standard form for the circle.