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December 19, 2014

December 19, 2014

Posted by **kyle** on Saturday, January 15, 2011 at 4:20pm.

- Math -
**Damon**, Saturday, January 15, 2011 at 5:07pm(x-k)^2+ (y-h)^2 = r^2

so

(1-k)^2 + (5-h)^2 = r^2

(3-k)^2 + (7-h)^2 = r^2

(5-k)^2 + (5-h)^2 = r^2

Now before I go off and solve those for k,h and r, sketch a graph

Notice that two points, A and C are at the same height,5

The third point,B, is half way between them (3 is halfway between 1 and 5)

I conclude that the center of the circle is on x = 3

so

k = 3

Onward

(1-3)^2 + (5-h)^2 = r^2

(3-3)^2 + (7-h)^2 = r^2

(5-3)^2 + (5-h)^2 = r^2

that second equation is pretty easy now.

(7-h)^2 = r^2

the first and third are actually the same now that we know what k isso use the first and the second

4 + (5-h)^2 = r^2

(7-h)^2 = r^2

29 - 10 h + h^2 = r^2

49 -14 h + h^2 = r^2

----------------------

-20 +4 h = 0

h = 5

well I guess you can take it from there

- Math -
**kyle**, Saturday, January 15, 2011 at 5:33pmThank you!

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