Posted by **Knights** on Tuesday, March 19, 2013 at 1:47pm.

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?

- analytic geometry/graphing problem -
**drwls**, Tuesday, March 19, 2013 at 2:12pm
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

- analytic geometry/graphing problem -
**Knights**, Wednesday, March 20, 2013 at 4:26pm
thanks a lot

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