Geometry HELP HELP
posted by Mary .
Find the equation of a circle circumscribes a triangle determined by the line y= 0 , y= x and 2x+3y= 10
PLEASE HELP ME BELLS

Make a sketch, and by some simple preliminary algebra you can see that the triangle is
A(0,0) , B(2,2) and C(0, 10/3)
the centre of the circle lies on the right bisectors of any of these sides
midpoint of AC = (0, 5/3) so rightbisector equation is
y = 5/3
midpoint of B(2,2) is (1,1)
slope of AB = 1
so slope of bisector at (1,1) is =1
y1 = 1(x1)
x + y = 2
sub in y = 5/3
x = 25/3 = 1/3
centre is (1/3 , 5/3)
equation:
(x  1/3)^2 + (y5/3)^2 = r^2
plug in (0,0) which lies on it
1/9 + 25/9 = r^2 = 26/9
(x1/3)^2 + (y5/3)^2 = 26/9 
oops, I misread the question, and got the wrong triangle
(I used the yaxis, instead of the xaxis)
so thepoints are (0,0), (2,2) and (5,0)
same equation for the rightbisector of AB which is
x + y = 2
the right bisector of line from A to (5,0) is
x = 5/2
for centre:
5/2 + y = 2
y = 1/2
centre is (5/2, 1/2)
(x5/2)^2 + (y+1/2)^2 = r^2
for (0,0) , which lies on the circle
25/4 + 1/4 = r^2 = 26/4
(x  5/2)^2 + (y+ 1/2)^2 = 26/4
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