How do you find the equation for a circle given center at (-1,-2) & passes through point (15,-2)?

Feb 1, 2016

${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = 256$

Explanation:

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 centre and $r$ is the radius.

This circle is therefore ${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = {r}^{2}$

We can use the given point to find $r$ but substitution.

${\left(15 + 1\right)}^{2} + {\left(- 2 + 2\right)}^{2} = {r}^{2}$

${r}^{2} = {16}^{2} = 0$
$\therefore r = 16$

The required equation is ${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = 256$