# How do you find the equation of a circle that passes through the points (1, 1), (1, 5), and (5, 5)?

Jun 25, 2016

${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$

#### Explanation:

Note that the centre of the circle is equidistant from all three points, the distance being the radius of the circle.

Any point equidistant from two points must lie on the perpendicular bisector of the line segment joining those two points. That is, on the line through the midpoint of the line segment, perpendicular to the line segment.

The perpendicular bisector of the line segment joining the points $\left(1 , 1\right)$ and $\left(1 , 5\right)$ is $y = 3$

The perpendicular bisector of the line segment joining the points $\left(1 , 5\right)$ and $\left(5 , 5\right)$ is $x = 3$

These intersect at the centre of the circle $\left(3 , 3\right)$

The distance between $\left(1 , 1\right)$ and $\left(3 , 3\right)$ is:

$\sqrt{{\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{8}$

So the equation of the circle may be written:

${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$

graph{((x-3)^2+(y-3)^2-8)((x-3)^2+(y-3)^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-1)^2+(y-5)^2-0.02)((x-5)^2+(y-5)^2-0.02)(y-3)(x-3+0.0001y)=0 [-7.625, 12.375, -2.72, 7.28]}