A triangle has corners at #(5 ,8 )#, #(2 ,7 )#, and #(7 ,3 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-03-05T17:33:18

#~~ 32.3#

Let the center of the circum-circle be #(h,k)# and its radius be #r#. Then, the equation of the circle is

#(x-h)^2+(y-k)^2 = r^2#

Since this circle must pass through the three vertices we have

#(5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2 = r^2#

From #(5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2# we get

#25-10h+h^2 +64-16k+k^2 = 4-4h+k^2+49-14k+k^2#

so that

#color(red)(6h + 2k = 36)#

Similarly, #(2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2# gives us

#4-4h+h^2 +49-14k+k^2 = 49-14h+h^2+9-6k+k^2 # which simplifies to

#color(red)(10h-8k = 5)#

Solving the two equations simultaneously gives

# h = 149/34,quad k = 165/34#

This leads to
#r^2 ~~ 1.29 #

so that the area of the circumscribed circle is
#pi r^2 ~~ 32.3#