# How do you find the general form of the equation of the circle with center (3,3) and containing the point (4,4)?

May 24, 2016

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

#### Explanation:

The general form for a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + \left(y - \textcolor{b l u e}{b}\right) = \textcolor{g r e e n}{{r}^{2}}$
where the center of the circle is at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
and the radius is $\textcolor{g r e e n}{r}$

If the center is at $\left(\textcolor{red}{3} , \textcolor{b l u e}{3}\right)$
and $\left(\textcolor{b r o w n}{4} , \textcolor{\mathmr{and} a n \ge}{4}\right)$ is a point on the circle,
then the radius is the distance from the center to (any) point on the circle
color(white)("XXX")color(green)(r)=sqrt((color(red)(3)-color(brown)(4))^2+(color(blue)(3)-color(orange)(4)^2)

$\textcolor{w h i t e}{\text{XXXX")=sqrt(2)color(white)("XX")rarrcolor(white)("XX}} \textcolor{g r e e n}{{r}^{2}} = \textcolor{g r e e n}{2}$

Therefore the equation of the circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{3}\right)}^{2} + {\left(y - \textcolor{b l u e}{3}\right)}^{2} = \textcolor{g r e e n}{2}$