A triangle has corners at (3,5), (7,9), and (4,2). What is the area of the triangle's circumscribed circle?

1 Answer
Douglas K.
Oct 12, 2016

A=145π8

Explanation:

Using the 3 points and the standard equation of a circle, r2=(x−h)2+(x−k)2, we can write 3 equations:

r2=(3−h)2+(5−k)2
r2=(7−h)2+(9−k)2
r2=(4−h)2+(2−k)2

Because r2=r2, we can set the right side of the first equation equal to the right side of the second equation and the third equation:

(3−h)2+(5−k)2=(7−h)2+(9−k)2
(3−h)2+(5−k)2=(4−h)2+(2−k)2

Expand the squares:

9−6h+h2+25−10k+k2=49−14h+h2+81−18k+k2
9−6h+h2+25−10k+k2=16−8h+h2+4−4k+k2

combine like terms and cancel the h2 and k2 terms:

8k=96−8h
−6k=−2h−14

Divide the first equation by 8:

k=12−h
−6k=−2h−14

Substitution:

−6(12−h)=−2h−14

−72+6h=−2h−14

8h=58

h=294

k=484−294

k=194

Substitute (294,194) for (h,k):

r2=(3−294)2+(5−194)2

r2=(−174)2+(14)2

r2=29016=1458

The area of a circle is πr2

A=145π8

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