# How do you find an equation of the circle that passes through the pts (0,0) and (1,1) and has its center on the line y=2?

Jan 20, 2016

#### Answer:

${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 5$

#### Explanation:

If the center of the circle is on $y = 2$
then the center of the circle is at $\left({x}_{c} , 2\right)$ for some value ${x}_{c}$

The radius of the circle, $r$, is

• The distance from $\left({x}_{c} , 2\right)$ to $\left(0 , 0\right)$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{\left({x}_{c} - 0\right)}^{2} + {\left(2 - 0\right)}^{2}} = \sqrt{{x}_{c}^{2} + 4}$
• The distance from $\left({x}_{c} , 2\right)$ to $\left(1 , 1\right)$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{\left({x}_{c} - 1\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{{x}_{c}^{2} - 2 {x}_{c} + 1 + 1}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {x}_{c}^{2} + 4 = {x}_{c}^{2} - 2 {x}_{c} + 2$

$\textcolor{w h i t e}{\text{XXX}} - 2 {x}_{c} = - 2$

$\textcolor{w h i t e}{\text{XXX}} {x}_{c} = - 1$

and since $r = \sqrt{{x}_{c}^{2} + 4}$
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{5}$

The general equation of a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

By substituting $\left(- 1 , 2\right)$ for $\left(a , b\right)$ and $\sqrt{5}$ for $r$, we get
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 5$
graph{(x+1)^2+(y-2)^2=5 [-5.8, 5.3, -0.764, 4.783]}