A triangle has corners at #(2 ,4 )#, #(3 ,6 )#, and #(4 ,7 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-21T13:11:41

# 65/2 pi #

Just did one just like this.

The circumscribed circle is just the circle with the three vertices on it. The general equation is

#(x-a)^2 + (y-b)^2 = r^2#

Substituting three points,

#(2-a)^2 + (4-b)^2=r^2#
#(3-a)^2+(6-b)^2=r^2#
#(4-a)^2+(7-b)^2=r^2#

Expanding,

# 20 = 4a + 8b + r^2-a^2-b^2 #
# 45 = 6a + 12 b + r^2-a^2-b^2 #
# 65 = 8a + 14b + r^2-a^2-b^2 #

Subtracting pairs,

# 25 = 2a + 4b #
#20 = 2a + 2b #

Subtracting,

#5 = 2b#
#b=5/2#
#a = 10 - b = 15/2#

Substituting,

#r^2 = (2-15/2)^2 + (4-5/2)^2= 65 /2 #

So the circle's area is

#A = \pi r^2 = 65/2 pi #