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

1 Answer
Dec 29, 2016

Answer:

The center is at #(4,1)#.enter image source here

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#).