# What are the coordinates of the center of the circle that passes through the points (1, 1), (1, 5), and (5, 5)?

Jul 30, 2016

$\left(3 , 3\right)$

#### Explanation:

Together with the point $\left(5 , 1\right)$ these points form the vertices of a $4 \times 4$ square with centre $\left(3 , 3\right)$, which is therefore the centre of the circle through the points too.

In general, given three points $A$, $B$ and $C$ we can construct perpendicular bisectors of line segments $A B$ and $B C$, then find their point of intersection, which will be the centre of a circle through them. This is because points on the perpendicular bisector of $A B$ are exactly those points which are equidistant from $A$ and $B$. So the intersection of the two perpendicular bisectors will be the unique point which is equidistant from $A$, $B$ and $C$.

In our example, let $A = \left(1 , 1\right)$, $B = \left(1 , 5\right)$ and $C = \left(5 , 5\right)$

Then the midpoint of $A B$ is $\left(1 , 3\right)$. Since $A B$ is vertical, its perpendicular bisector is horizontal through $\left(1 , 3\right)$, i.e. the line $y = 3$

The midpoint of $B C$ is $\left(3 , 5\right)$. Since $B C$ is horizontal, its perpendicular bisector is vertical through $\left(3 , 5\right)$, i.e. the line $x = 3$.

The lines $y = 3$ and $x = 3$ intersect at the point $\left(3 , 3\right)$ as expected.

graph{((x-3)^100+(y-3)^100-2^100)(y-3)(x-3)((x-3)^2+(y-3)^2-8)((x-3)^2+(y-3)^2-0.024)((x-1)^2+(y-5)^2-0.02)((x-5)^2+(y-5)^2-0.02)((x-1)^2+(y-1)^2-0.02)=0 [-6.75, 13.25, -2.88, 7.12]}