Question

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

Answer 1

The center is at `(4,1)`.

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 `(0,-2)` and `(8,-2)` therefore its perpendicular bisector is `x=(0+8)/2`, that is, `x=4`.

Another side joins `(8,-2)` and `(7,-3)` so its mid point is `((15/2),-5/2)` and its slope is `(-3-(-2))/(7-8)` = `1`. Therefore the perpendicular bisector has slope `-1/1` = `-1` and its equation is `y+x=15/2-5/2=5`.

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

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