# Find the equation of circle which touches both the negative axes and has its center on the line x-2y=3.?

Oct 23, 2017

${\left(x + 3\right)}^{2} + {\left(y + 3\right)}^{2} = 9$

#### Explanation:

The center of the circle must have equal $x$ and $y$ coordinate values if the circle touches both axes.

For the line $x - 2 y = 3$
this means that (after substituting $x$ for $y$, since they need to be equal)
$\textcolor{w h i t e}{\text{XXX")x-2x=3color(white)("xxx")rarrcolor(white)("xxx}} x = - 3$
and similarly
$\textcolor{w h i t e}{\text{XXX}} y = - 3$

With a center at $\left(\textcolor{g r e e n}{- 3} , \textcolor{b l u e}{- 3}\right)$ and a radius of $\textcolor{m a \ge n t a}{3}$ (since this is the distance from the center to both axes),
the equation of the circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(\textcolor{g r e e n}{- 3}\right)\right)}^{2} + {\left(y - \left(\textcolor{b l u e}{- 3}\right)\right)}^{2} = {\textcolor{red}{3}}^{2}$
or
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 3\right)}^{2} + {\left(y + 3\right)}^{2} = 9$