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

Jan 2, 2016

${\left(x - \frac{7}{2}\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(\frac{5}{2}\right)}^{2}$

#### Explanation:

First plot the points out.

Let $D$ be the midpoint of the line segment $B C$. The position of $D$ is $\left(2 , 1\right)$. 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 $A C E$ and angle $A B E$ are both right angles. Using similar triangles we get the position of $E$ to be $\left(6 , 1\right)$. 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 $A E$ is a diameter of the circle.

The radius is $\frac{5}{2}$ and the center is $\left(\frac{7}{2} , 1\right)$.

The equation of the circle is:

${\left(x - \frac{7}{2}\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(\frac{5}{2}\right)}^{2}$