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)#. enter image source here

The point where the perpendicular bisector of the triangle's three sides intersects is where the center is located.

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#).

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Answer 2

Benjamin Achtenberg

To find the center of the circle circumscribed about a triangle, you need to find the intersection point of the perpendicular bisectors of any two sides of the triangle.

Find the midpoints of any two sides of the triangle.
Find the slopes of these two sides.
Find the negative reciprocal of these slopes to get the slopes of the perpendicular bisectors.
Use the midpoint and the slope of each side to write the equations of the perpendicular bisectors.
Solve these equations simultaneously to find the point of intersection, which is the center of the circumscribed circle.

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