Draw a chord and the its perpendicular bisector.
Draw a 2nd chord and its perpendicular bisector.
The two perpendicular bisectors will intersect at the center of the circle.
2007-10-04 05:05:23
·
answer #1
·
answered by ironduke8159 7
·
0⤊
0⤋
It depends on what values you have to work with.
You may be able to work it out using the equation of a circle.
(x-a)^2+(y-b)^2=r^2 where (a,b) are the coordinates of the center of the circle and r is the radius.
2007-10-04 05:13:20
·
answer #2
·
answered by Atari 2
·
0⤊
0⤋
Geometrically, you can find the center of a circle by intersecting the perpendicular bisectors of three points on the circle.
For coordinates, if you are given three points on the circle, P1(x1,y1), P2(x2,y2), and P3 (x3,y3), then the center C(x,y) is equidistant from the three points on the circle.
distance = the square root of [(x2-x1)^2 + (y2-y1)^2]
2007-10-04 05:06:45
·
answer #3
·
answered by Hiker 4
·
0⤊
0⤋
draw a tangent to the circle and a perpendicular to that tangent. The center will be on that perpendicular half way across.
2007-10-04 05:05:57
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
Divide the perimeter in half
2007-10-04 05:07:06
·
answer #5
·
answered by Anonymous
·
0⤊
0⤋