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December 7, 2016
Posted by **Mary ** on Tuesday, February 18, 2014 at 6:56am.

PLEASE HELP ME BELLS

- Geometry HELP HELP -
**Reiny**, Tuesday, February 18, 2014 at 12:30pmMake 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 - back up - Geometry HELP HELP -
**Reiny**, Tuesday, February 18, 2014 at 12:37pmoops, 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