Posted by Mary on Tuesday, February 18, 2014 at 6:56am.
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 right-bisector 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
y-1 = -1(x-1)
x + y = 2
sub in y = 5/3
x = 2-5/3 = 1/3
centre is (1/3 , 5/3)
equation:
(x - 1/3)^2 + (y-5/3)^2 = r^2
plug in (0,0) which lies on it
1/9 + 25/9 = r^2 = 26/9
(x-1/3)^2 + (y-5/3)^2 = 26/9
oops, I misread the question, and got the wrong triangle
(I used the y-axis, instead of the x-axis)
so thepoints are (0,0), (2,2) and (5,0)
same equation for the right-bisector 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)
(x-5/2)^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