# The point (-3,2) lies on a circumference whose equation is (x+3)^2+(y+1)^2-r^2 = 0, what is the radius of the circumference?

Dec 25, 2016

The radius is $r = 3$.

#### Explanation:

If the point $\left(- 3 , 2\right)$ belongs to the circle ${\left(x + 3\right)}^{2} + {\left(y + 1\right)}^{2} - {r}^{2} = 0$ then its coordinates must verify that equation. That is, if we substitute in the equation of the circle $x$ by $- 3$ and $y$ by $2$ we must obtain a true equality. Then, it is enough to clear $r$ of that equation:

${\left(- 3 + 3\right)}^{2} + {\left(2 + 1\right)}^{2} - {r}^{2} = 0 \Rightarrow 9 - {r}^{2} = 0$,

then:

$r = \sqrt{9} = 3$.

(Remember $r$ is a distance, then it's always a positive number).