analytic geometry/graphing problem
posted by Knights on .
The vertices of a triangle are the points of intersection of the line y=-x-1, 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?
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
-x-1 = 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
thanks a lot