# What is the center of a circle circumscribed about a triangle with vertical (-2,2) (2,-2) (6,-2)?

Jun 4, 2016

$\left(4 , 4\right)$

#### Explanation:

The centre of a circle passing through two points is equidistant from those two points. Therefore it lies on a line which passes through the midpoint of the two points, perpendicular to the line segment joining the two points. This is called the perpendicular bisector of the line segment joining the two points.

If a circle passes through more than two points then its centre is the intersection of the perpendicular bisectors of any two pairs of points.

The perpendicular bisector of the line segment joining $\left(- 2 , 2\right)$ and $\left(2 , - 2\right)$ is $y = x$

The perpendicular bisector of the line segment joining $\left(2 , - 2\right)$ and $\left(6 , - 2\right)$ is $x = 4$

These intersect at $\left(4 , 4\right)$

graph{(x-4+y*0.0001)(y-x)((x+2)^2+(y-2)^2-0.02)((x-2)^2+(y+2)^2-0.02)((x-6)^2+(y+2)^2 - 0.02)((x-4)^2+(y-4)^2-40)((x-4)^2+(y-4)^2-0.02) = 0 [-9.32, 15.99, -3.31, 9.35]}

Jun 4, 2016

(4, 4)

#### Explanation:

Let the center be C(a, b)..

As the vertices are equidistant from the center,

${\left(a + 2\right)}^{2} + {\left(b - 2\right)}^{2} = {\left(a - 2\right)}^{2} + {\left(b + 2\right)}^{2} = {\left(a - 6\right)}^{2} + {\left(b + 2\right)}^{2}$

Subtracting 2nd from the first and the third from the second,

a - b = 0 and a = 4. So, b = 4.

So, the center is C(4, 4).
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