Question

The point (-3,2) lies on a circumference whose equation is `(x+3)^2+(y+1)^2-r^2 = 0`, what is the radius of the circumference?

Answer 1

The radius is `r = 3`.

Explanation:

If the point `(-3, 2)` belongs to the circle `(x + 3)^2 + (y + 1)^2 - r^2 = 0` then its coordinates must verify that equation. That is, if we substitute in the equation of the circle `x` by `- 3` and `y` by `2` we must obtain a true equality. Then, it is enough to clear `r` of that equation:

`(-3 + 3)^2 + (2 + 1)^2 - r^2 = 0 rArr 9 - r^2 = 0`,

then:

`r = sqrt 9 = 3`.

(Remember `r` is a distance, then it's always a positive number).