Question

How do you graph the circle with center at (-1, 2) and diameter 12?

Answer 1

Draw a circle with center at `(-1, 2)` and radius of `6`. Equation of the circle is `x^2+y^2+2x-2y-31=0`

Explanation:

To graph the circle with center at `(-1, 2)` and diameter `12` means a radius of `12/2=6`, just draw a circle with center at `(-1, 2)` and radius of `12/2=6`.

As a point on circle is always a a distance of `6` from `(-1,2)`

`(x-(-1))^2+(y-2)^2=36` or `(x+1)^2+(y-2)^2=36` i.e.

`x^2+2x+1+y^2-2y+4=36` or

`x^2+y^2+2x-2y-31=0`

graph{x^2+y^2+2x-2y-31=0 [-15.8, 15.81, -7.9, 7.9]}