# What is a equation that has a center at (-5,2) with a radius of 6?

Apr 6, 2018

The given point (center of the circle) is $\left(- 5 , 2\right)$
Let the unknown point on the circumference of the circle be $\left({x}_{1} , {y}_{1}\right)$

color(white)(dd

The radius of the circle is given by $\sqrt{{\left({x}_{1} - h\right)}^{2} + {\left({y}_{1} - k\right)}^{2}}$
(its just basically the distance formula)

color(white)(dd

And therefore, the general equation of a circle having a center $\left(h , k\right)$ and radius $r$ is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$.

So, in this case, since the center is $\left(- 5 , 2\right)$
${\left(x - \left(- 5\right)\right)}^{2} + {\left(y - \left(2\right)\right)}^{2} = {\left(6\right)}^{2}$.

$\implies {\left(x + 5\right)}^{2} + {\left(y - 2\right)}^{2} = 36$.

$\implies {x}^{2} + 25 + 10 x + {y}^{2} + 4 + 4 x = 36$

=> color(red)(x^2 + y^2 + 10x + 4y - 7=0

Apr 6, 2018

${\left(x + 5\right)}^{2} + {\left(y - 2\right)}^{2} = 36$

#### Explanation:

$\text{the standard form of the equation of a circle is}$

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

$\text{where "(a,b)" are the coordinates of the centre and r is}$
$\text{the radius}$

$\text{here "(a,b)=(-5,2)" and } r = 6$

${\left(x - \left(- 5\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {6}^{2}$

$\Rightarrow {\left(x + 5\right)}^{2} + {\left(y - 2\right)}^{2} = 36 \leftarrow \textcolor{red}{\text{equation of circle}}$