# How do you write the equation for a circle with center at (2,-5) and passing through (3,4)?

Jun 10, 2016

${\left(x - 2\right)}^{2} + {\left(y + 5\right)}^{2} = 21$

#### Explanation:

The general equation of a circle is:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
$x =$x-coordinate
$h =$x-coordinate of centre
$y =$y-coordinate
$k =$y-coordinate of centre
$r =$radius

Start by plugging $\left(h , k\right) = \left(2 , - 5\right)$ into the equation.

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

Replace $\left(x , y\right)$ with $\left(3 , 4\right)$.

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

Simplify.

${\left(1\right)}^{2} + {\left(9\right)}^{2} = {r}^{2}$

$1 + 81 = {r}^{2}$

$21 = {r}^{2}$

Rewrite the equation by including the centre and the radius.

${\left(x - 2\right)}^{2} + {\left(y - \left(- 5\right)\right)}^{2} = 21$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - 2\right)}^{2} + {\left(y + 5\right)}^{2} = 21} \textcolor{w h i t e}{\frac{a}{a}} |}}}$