What is the general equation of a circle which passes through #A#(1,1), #B#(2,-1),and #C#(2,3)?

1 Answer
Jan 2, 2016

Answer:

#(x-7/2)^2 + (y-1)^2 = (5/2)^2#

Explanation:

First plot the points out.

Let #D# be the midpoint of the line segment #BC#. The position of #D# is #(2,1)#. Note that ADC and ABC are right angle triangles.

An important concept is that a triangle formed by joining the two ends of a diameter and any other point on the circumference will always be a right angle triangle.

We want to find a diameter on our circle. Therefore we want to construct a right angle triangle with all 3 points on the circumference of the circle.

To do so, we observe that there must be a point, #E#, such that angle #ACE# and angle #ABE# are both right angles. Using similar triangles we get the position of #E# to be #(6,1)#. Triangle ADC is similar to triangle CDE. Triangle ADB is similar to triangle BDE. The linear scaling factor is 2.

Try to understand why point #E# must lie on the circumference of the circle.

Now we know that #AE# is a diameter of the circle.

The radius is #5/2# and the center is #(7/2,1)#.

The equation of the circle is:

#(x-7/2)^2 + (y-1)^2 = (5/2)^2#