# How do you find the center of the circle that is circumscribed about the triangle with vertices (0,-2), (7,-3) and (8,-2)?

Dec 29, 2016

The center is at $\left(4 , 1\right)$. #### Explanation:

The center lies at the point of intersection of the perpendicular bisector of the three sides of the triangle.

One of these sides is joins $\left(0 , - 2\right)$ and $\left(8 , - 2\right)$ therefore its perpendicular bisector is $x = \frac{0 + 8}{2}$, that is, $x = 4$.

Another side joins $\left(8 , - 2\right)$ and $\left(7 , - 3\right)$ so its mid point is $\left(\left(\frac{15}{2}\right) , - \frac{5}{2}\right)$ and its slope is $\frac{- 3 - \left(- 2\right)}{7 - 8}$ = $1$. Therefore the perpendicular bisector has slope $- \frac{1}{1}$ = $- 1$ and its equation is $y + x = \frac{15}{2} - \frac{5}{2} = 5$.

The center, $\left(a , b\right)$ lies on both these lines, therefore $a = 4$ and $b = 1$. So the center is at $\left(4 , 1\right)$.

Alternatively, substitute the co-ordinates of the three points into the standard equation ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ and solve the three simultaneous equations for $a$, $b$ (and $r$, which comes out as $5$).