Posted by Knights on .
The vertices of a triangle are the points of intersection of the line y=x1, x=2 and y = 1/5x + 13/5. Find an equation of the circle passing through all three vertices.
I don't understand how to solve this: should I set them all equal to find the vertices? But afterwards, then what?

analytic geometry/graphing problem 
drwls,
First get the coordinates of the three vertices. It will help to plot the points.
The first two lines meet at x=2,y=3
The second pair of lines meet at x=2, y = 2/5 +13/5 = 3
The first and third lines meet at
x1 = x/5 + 13/5
6x/5 = 18/5
x = 3, y = 2
The center of the circle through these points is located where two perpendicular side bisectors meet. Two side bisectors are y=0 and x = y. They meet at the origin: (0,0)
The radius of the circle is the distance from the origin to any vertex. That would be sqrt(4+9) = sqrt13
So the circle equation that passes through the three vertices is
x^2 + y^2 = 13 
analytic geometry/graphing problem 
Knights,
thanks a lot