# How do you find the equation in general form for the circle centered at (-3,4) and passing through the origin?

May 9, 2018

${x}^{2} + 6 x + {y}^{2} - 8 y = 0$

#### Explanation:

Given: circle centered at $\left(- 3 , 4\right)$ and passing through $\left(0 , 0\right)$

The standard form of a circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where $\left(h , k\right)$ is the center and $r =$ radius

Since the circle passes through $\left(0 , 0\right)$, the radius can be calculated by find the distance between the center and the origin:

$r = \sqrt{{\left(- 3\right)}^{2} + {4}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

The circle in standard form is: ${\left(x + 3\right)}^{2} + {\left(y - 4\right)}^{2} = 25$

To put the circle in general form you need to distribute.

Use the square rule: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

${x}^{2} + 6 x + 9 + {y}^{2} - 8 y + 16 - 25 = 0$

${x}^{2} + 6 x + {y}^{2} - 8 y = 0$