Geometry of the Plane

Circles

The formula for a circle with center $(h,k)$ and radius $r$ is $$ (x-h)^2 + (y-k)^2 = r^2. $$

Example.

Find the equation of a circle that passes through the points $(1,3)$, $(1,5)$, and $(3,5)$.

We plug in these values in the formula as the $x$ and $y$ values: \begin{align} (1-h)^2 + (3-k)^2 & = r^2 \\ (1-h)^2 + (5-k)^2 & = r^2 \\ (3-h)^2 + (5-k)^2 & = r^2 \end{align} We now solve this (nonlinear) system of 3 equations to 3 unknowns.
Setting the two first equations equal to each other: $$ (3-k)^2 = (5-k)^2, $$ which implies that $k=4$. (Details) So $$ 3-k = \pm (5-k) $$ which implies that either $$ 3-k = 5-k \; , \quad \text{thus } \; 3=5 $$ (which is FALSE), or $$ 3-k = -(5-k) \; , \quad \text{so } \; k=4. $$ We deduce that $k=4$.

Setting the last two equations equal to each other : $$ (1-h)^2 = (3-h)^2, $$ which implies that $h=2$. (Details) So $$ 1-h = \pm (3-h) $$ which implies that either $$ 1-h = 3-h \; , \quad \text{thus } \; 1=3 $$ (which is FALSE), or $$ 1-h = -(3-h) \; , \quad \text{so } \; h=2. $$ We deduce that $h=2$.

We already know the center of the circle: $(2,4)$

To find the radius, we can just use one of the equations: $$ (1-2)^2 + (3-4)^2 = r^2 \quad so \quad r^2 = 2, $$ and we have $r = \sqrt{2}$.

The equation of the circle that passes through these 3 points is: $$ (x-2)^2+(y-4)^2=2. $$