Question

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

Answer 1

`x^2+y^2=40`

See below.

Explanation:

The equation of a circle with center `(h,k)` 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. `(0,0)`, we can use this fact along with the point which the circle passes through `(-6,-2)` to find the radius. This is done using the distance formula.

`r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Given the points `(0,0)` and `(-6,-2)`:

`r=sqrt((-6-0)^2+(-2-0)^2)`

`=sqrt(36+4)`

`sqrt(40)=2sqrt(10)`

`=>r=2sqrt(10)`.

Therefore, the equation of the circle is given by:

`x^2+y^2=(2sqrt(10))^2`

`=>x^2+y^2=40`

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