# How do you find the equation given center passes through the origin and has its center at (0,4)?

Dec 23, 2016

${x}^{2} + {\left(y - 4\right)}^{2} = 16$

#### Explanation:

Assuming this is to be the equation of a circle (although it was asked under the more general "Geometry of an Ellipse")

A circle with center $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and radius $\textcolor{g r e e n}{r}$ has the equation:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$

If the required circle has its center at $\left(0 , 4\right)$
then $\textcolor{red}{a} = \textcolor{red}{0}$
and $\textcolor{b l u e}{b} = \textcolor{b l u e}{4}$

Further if its center is at $\left(0 , 4\right)$ and it passes through the origin i.e. through $\left(0 , 0\right)$ it radius must be $\textcolor{g r e e n}{r} = \textcolor{g r e e n}{4}$

So the equation of the required circle must be
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{0}\right)}^{2} + {\left(y - \textcolor{b l u e}{4}\right)}^{2} = {\textcolor{g r e e n}{4}}^{2}$
or, simplifying slightly:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(y - 4\right)}^{2} = 16$