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Hi Lindsay. We know that any three points (as long as they aren't on a single line) define a circle, so when we mark these coordinates on a graph, we can draw the circle: The equation of any circle is (xa)^{2}+(yb)^{2}=r^{2}, where (a,b) is the center and r is the radius. You have three values for (x, y) on the circumference of the circle, so that can create three equations with three unknown values (a, b, and r):
So three equations with three unknown values can be solved in the same way as simultaneous equations are normally solved: with the substitution and/or the elimination methods. Solving equations (1) and (2) for b^{2} gives us:
from which you can eliminate r^{2} from both sides and find a. Once you know a, you can find b and r the same way. Then you will be able to put your values of a, b and r into the equation of the circle(xa)^{2}+(yb)^{2}=r^{2}. Cheers,
Lindsay, I thought of an even easier way to solve this problem. The center of the circle has to be halfway between the y intercepts and halfway between the x intercepts. That immediately tells you the coordinates of the center. Then you can use the distance formula to any of the three intercepts (the origin would be easiest) to find the radius. The method I used (and which Penny uses) will work for any three points, not just intercepts, so it is more powerful, but this shortcut works for intercepts. Stephen.
 


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