# How do you write the standard from of the equation of the circle given center C ( 6, - 2 ), radius 2?

Jul 23, 2018

${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 4$

#### Explanation:

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

•color(white)(x)(x-a)^2+(y-b)^2=r^2

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

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

${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 4 \leftarrow \textcolor{b l u e}{\text{equation of circle}}$

Jul 23, 2018

${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 4$

#### Explanation:

Recall that the equation of a circle is given by

$\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \textcolor{w h i t e}{\frac{2}{2}} |}}$, with center $\left(h , k\right)$ and radius $r$

We know that we are centered at $\left(6 , - 2\right)$ and have a radius of $2$, so $h = 6 , k = - 2$ and ${r}^{2} = 4$.

We can plug these into our equation to get

${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 4$

Hope this helps!