# How do you write the general form given a circle that passes through the given (x-2)^2 + (y+1)^2 = 9?

Apr 1, 2016

${x}^{2} + {y}^{2} - 4 x + 2 y - 4 = 0$

#### Explanation:

The general form of the equation of a circle is.

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

To write the given equation in this form , requires expanding the brackets and rearranging into the general form.

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

hence: ${x}^{2} - 4 x + 4 + {y}^{2} + 2 y + 1 = 9$

$\Rightarrow {x}^{2} + {y}^{2} + 4 x + 2 y - 4 = 0 \text{ is in general form }$