How do you find the equation of a circle with center at the origin and passing through (-6,-2)?

Mar 2, 2017

${x}^{2} + {y}^{2} = 40$

See below.

Explanation:

The equation of a circle with center $\left(h , k\right)$ and radius $r$ is given by (x−h)^2+(y−k)^2=r^2. For a circle centered at the origin, this becomes the more familiar equation ${x}^{2} + {y}^{2} = {r}^{2}$.

Because we know the circle is centered at the origin, i.e. $\left(0 , 0\right)$, we can use this fact along with the point which the circle passes through $\left(- 6 , - 2\right)$ to find the radius. This is done using the distance formula.

$r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Given the points $\left(0 , 0\right)$ and $\left(- 6 , - 2\right)$:

$r = \sqrt{{\left(- 6 - 0\right)}^{2} + {\left(- 2 - 0\right)}^{2}}$

$= \sqrt{36 + 4}$

$\sqrt{40} = 2 \sqrt{10}$

$\implies r = 2 \sqrt{10}$.

Therefore, the equation of the circle is given by:

${x}^{2} + {y}^{2} = {\left(2 \sqrt{10}\right)}^{2}$

$\implies {x}^{2} + {y}^{2} = 40$

graph{x^2+y^2=40 [-20, 20, -10, 10]}