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?

1 Answer
Pitufox27
Dec 25, 2016

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).

Related questions
See all questions in Equation of a Circle
Impact of this question
6727 views around the world
You can reuse this answer
Creative Commons License