The point (-4,-3) lies on a circle whose center is at (0,6). How do you find an equation of this circle?

Feb 15, 2016

${x}^{2} + {\left(y - 6\right)}^{2} = 109$

Explanation:

If the circle has a center at $\left(0 , 6\right)$ and $\left(- 4 , - 3\right)$ is a point on its circumference,
then it has a radius of:
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(0 - \left(- 3\right)\right)}^{2} + {\left(6 - \left(- 4\right)\right)}^{2}} = \sqrt{109}$

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

In this case we have
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(y - 6\right)}^{2} = 109$
graph{x^2+(y-6)^2=109 [-14.24, 14.23, -7.12, 7.11]}

Feb 15, 2016

${x}^{2} + {y}^{2} + 8 x + 6 y - 72 = 0$

Explanation:

It means that $\left(- 4 , - 3\right)$ is center and radius is distance between $\left(- 4 , - 3\right)$ and $\left(0 , 6\right)$. The radius is hence given by

$\sqrt{{\left(0 - \left(- 4\right)\right)}^{2} + {\left(6 - \left(- 3\right)\right)}^{2}}$ or $\sqrt{16 + 81}$ or $\sqrt{87}$

Hence equation of circle is

${\left(x - \left(- 4\right)\right)}^{2} + \left(y - \left(- {3}^{2}\right)\right) = 87$ or

${\left(x + 4\right)}^{2} + {\left(y + 3\right)}^{2} = 87$

${x}^{2} + 8 x + 16 + {y}^{2} + 6 y + 9 = 87$ or

${x}^{2} + {y}^{2} + 8 x + 6 y + 16 + 9 - 87 = 0$ or

${x}^{2} + {y}^{2} + 8 x + 6 y - 72 = 0$