How do you find the area of a circle given 3 points on the coordinate plane?

Answer 1

Doug's on a roll - my sentiments exactly

Answer 2

using basic geometry. I would take the equation of a circle: (x-h)^2 + (y-k)^2 = r^2, FOIL it out and get x^2-2xh+h^2+y^2-2yk+k^2=r^2. I would then rearrange it like this: x^2-2xh+y^2-2yk = r^2-h^2-k^2. I would then plug in the 3 different points they gave you to get 3 separate equations. The right hand side is going to be the same in all 3 equations since the radius and center isn't changing, so you can set the left sides equal to each other and solve for h,k. Once you have the center, you can plug back in to find the radius. Then it's simply A=(pi)r^2
There are other ways, but I'm not sure what level math you're doing.

Answer 3

Develop the equations for straight line segments between any two of the three points given. Now develop the equations for perpendicular bisectors for each of those lines. Where the perpendicular bisectors cross (the common solution of the two equations) is the center of the circle. The distance from the center to any of the points is found using Pythagorean Theorem and is also the radius of the circle. Then it's just πr² and you're done.


Doug

Answer 4

Draw two lines, one between the first two points and the other between the last two. Draw perpendicular bisectors to each of these. The point of intersection of the bisectors is the center of the circle. Distance from the center to any of the points is the radius, and you can determine the area from that.

Answer 5

say points and coordinates are (these coordinates are known):
p1(x1,y1)
p2(x2,y2)
p3(x3,y3)
and center of circle C(xc,yc) (these coordinates are not known)
radius = r

this may help

(xc-x1)^2+(yc-y1)^2 = (xc-x2)^2+(yc-y2)^2 = (xc-x3)^2+(yc-y3)^2 = r^2

area = pi*r*r

Answer 6

jjnk