# What is the length of the radius and the coordinates of the center of the circle defined by the equation (x+7)^2+(y-3)^2=121?

Jul 6, 2017

In

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

The radius is given by $r$ and the centre by $\left(p , q\right)$. Hence, our centre is $\left(- 7 , 3\right)$ and the radius is $11$. The graph of the relation confirms.

graph{(x+ 7)^2 + (y - 3)^2 = 121 [-10, 10, -5, 5]}

Hopefully this helps!

Jul 6, 2017

Radius is $11$ and coordinates of center are $\left(- 7 , 3\right)$

#### Explanation:

Equation of a circle whose center is $\left(h , k\right)$ and radius is $r$ is

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

As the equation ${\left(x + 7\right)}^{2} + {\left(y - 3\right)}^{2} = 121$ can be written as

${\left(x - \left(- 7\right)\right)}^{2} + {\left(y - 3\right)}^{2} = {11}^{2}$

Radius is $11$ and coordinates of center are $\left(- 7 , 3\right)$