(3) In geometry, it is shown that three points in the plane, not in a straight line, determine a unique circle which passes through those three points.
(a) Find the equation of the circle which passes through (0,0), (0,1)
and (2,0).
(b) Describe a procedure that will always work to find the circle which
passes through three given noncollinear points.
SOLUTION:
(a) (x - X)2 + (y - Y)2 = R2
* At (0,0) the equation becomes X2 + Y2 = R2
==> R2 - Y2 = X2
At (0,1) the equation becomes X2 + 1 - 2Y + Y2 = R2
==> X2 + Y2 = X2 + 1 - 2Y + Y2
0 = 1 - 2Y ==> Y = 1/2
At (2,0) the equation becomes (2 - X)2 + Y2 = R2
==> R2 - Y2 = 4 - 4X + X2
From * we then get x2 = 4 - 4X + X2
==> 0 = 4 - 4X ==> X = 1
Since X2 + Y2 = R2
==> R2 = 1 + 1/4 = 5/4
Therefore we get:
(x - 1)2 + (y - 1/2)2 = 5/4
(b) Substituting in the points into the general equation.