A triangle has corners at (9 ,1 ), (4 ,6 ), and (7 ,4 ). What is the area of the triangle's circumscribed circle?

1 Answer
Douglas K.
Dec 24, 2017

"Area" = (169pi)/2

Explanation:

One way to solve this problem is as follows.

The vertices of the triangle are, also, points on the circumscribed circle; this allow us to use the Cartesian equation for a circle, (x-h)^2+(y-k)^2=r^2, and the 3 given (x,y) points, (9,1), (4,6), and (7,4) to write 3 equations:

(9-h)^2+(1-k)^2=r^2" [1]"
(4-h)^2+(6-k)^2=r^2" [2]"
(7-h)^2+(4-k)^2=r^2" [3]"

where (h,k) is the center of the circumscribed circle and r is its radius.

After a lot of non-linear algebra, you can verify that r = 13/sqrt2

The formula for the area of the circumscribed circle is:

"Area" = pir^2

Substitute r = 13/sqrt2:

"Area" = (169pi)/2

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