Question

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

Answer 1

`(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`