# How do you graph the circle with center at (-1, 2) and diameter 12?

Feb 24, 2016

Draw a circle with center at $\left(- 1 , 2\right)$ and radius of $6$. Equation of the circle is ${x}^{2} + {y}^{2} + 2 x - 2 y - 31 = 0$

#### Explanation:

To graph the circle with center at $\left(- 1 , 2\right)$ and diameter $12$ means a radius of $\frac{12}{2} = 6$, just draw a circle with center at $\left(- 1 , 2\right)$ and radius of $\frac{12}{2} = 6$.

As a point on circle is always a a distance of $6$ from $\left(- 1 , 2\right)$

${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 2\right)}^{2} = 36$ or ${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 36$ i.e.

${x}^{2} + 2 x + 1 + {y}^{2} - 2 y + 4 = 36$ or

${x}^{2} + {y}^{2} + 2 x - 2 y - 31 = 0$

graph{x^2+y^2+2x-2y-31=0 [-15.8, 15.81, -7.9, 7.9]}