# How do you find the equation of the circle given center (-1,4) and radius 3?

Oct 30, 2016

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

#### Explanation:

If the center of circle is $\left(h , k\right)$ and radius is $r$, its equation is

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

Hence equation of circle with center at $\left(- 1 , 4\right)$ and radius $3$, is

${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 4\right)}^{2} = {3}^{2}$

or ${\left(x + 1\right)}^{2} + {\left(y - 4\right)}^{2} = {3}^{2}$

or ${x}^{2} + 2 x + 1 + {y}^{2} - 8 y + 16 = 9$

or ${x}^{2} + {y}^{2} + 2 x - 8 y + 1 + 16 - 9 = 0$

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