# How do you write the equation of the circle with the given center at (-2,4) and passes through the point at (1,-7)?

Mar 2, 2017

Equation of the circle is ${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 121$ or ${x}^{2} + {y}^{2} + 4 x - 8 y - 101 = 0$

#### Explanation:

Equation of a circle with center at $\left(h , k\right)$ is

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, where $r$ is its radius.

Hence equation of a circle with center at $\left(- 2 , 4\right)$ is

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

As it passes through $\left(1 , - 7\right)$, we will have

${\left(- 2 + 2\right)}^{2} + {\left(- 7 - 4\right)}^{2} = {r}^{2}$

or ${0}^{2} + {11}^{2} = {r}^{2}$ and ${r}^{2} = 0 + 121 = 121$

Hence, equation of circle is ${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 121$

or ${x}^{2} + {y}^{2} + 4 x - 8 y - 101 = 0$ and its radius is $\sqrt{121} = 11$
graph{x^2+y^2+4x-8y-101=0 [-28, 24, -10, 16]}