How do you write the standard from of the equation of the circle given center (-2, 4) and radius 7?

Sep 14, 2016

${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 49$

Explanation:

The standard or general 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 of the circle and $r$ is the radius.

In this problem, $h = - 2$, $k = 4$ and $r = 7$

Plugging these values into the equation gives:

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

${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 49$

Some teachers refer to the "standard form" as the equation resulting from squaring the binomials and gathering all terms on one side. This is uncommon, but here it is.

$\left(x + 2\right) \left(x + 2\right) + \left(y - 4\right) \left(y - 4\right) = 49$

${x}^{2} + 4 x + 4 + {y}^{2} - 8 y + 16 = 49$

Rearranging and combining the constant terms gives

${x}^{2} + {y}^{2} + 4 x - 8 y - 29 = 0$