# How do you find the equation of a circle passes through point (-3,6) & touch the two axis?

Jul 19, 2016

Two solutions:

${\left(x + 15\right)}^{2} + {\left(y - 15\right)}^{2} = {15}^{2}$

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

#### Explanation:

The centre of the circle is in Q2 since the point $\left(- 3 , 6\right)$ is.

So the equation of the circle may be written:

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

This is satisfied by $\left(- 3 , 6\right)$, so:

${\left(- 3 + r\right)}^{2} + {\left(6 - r\right)}^{2} = {r}^{2}$

Expanding:

$9 - 6 r + {r}^{2} + 36 - 12 r + {r}^{2} = {r}^{2}$

Simplifying:

$0 = {r}^{2} - 18 r + 45$

$= {r}^{2} - 18 r + 81 - 36$

$= {\left(r - 9\right)}^{2} - {6}^{2}$

$= \left(\left(r - 9\right) - 6\right) \left(\left(r - 9\right) + 6\right)$

$= \left(r - 15\right) \left(r - 3\right)$

So $r = 15$ or $r = 3$

graph{((x+15)^2+(y-15)^2-15^2)((x+3)^2+(y-3)^2-3^2)((x+3)^2+(y-6)^2-0.04) = 0 [-16.51, 8.8, -1.37, 11.29]}