A triangle has corners at #(9 ,8 )#, #(2 ,3 )#, and #(7 ,4 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-09-28T16:04:18

#"Area" = 2405/81pi#

The standard Cartesian form for the equation of a circle is:

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

Where #(h,k)# is the center point and #r# is the radius.

We can use the equation and the 3 points to write 3 equations:

#(9-h)^2+(8-k)^2=r^2" [1]"#
#(2-h)^2+(3-k)^2=r^2" [2]"#
#(7-h)^2+(4-k)^2=r^2" [3]"#

Expand the squares:

#81-18h+h^2+64-16k+k^2=r^2" [1.1]"#
#4-4h+h^2+9-6k+k^2=r^2" [2.1]"#
#49-14h+h^2+16-8k+k^2=r^2" [3.1]"#

Subtract equation [2.1] from equation [1.1]:

#-14h-10k+ 132 = 0#

Write in standard form:

#7h + 5k = 66" [4]"#

Subtract equation [2.1] from equation [3.1]:

#-10h -2k + 52 = 0#

Write as k in terms of h:

# -2k =-52 +10h#

#k = 26 -5h" [5]"#

Substitute equation [5] into equation [4]:

#7h + 5(26 -5h) = 66#

#7h + 130 - 25h = 66#

#-18h=-64#

#h = 32/9#

Use equation [5] to find the value of k:

#k = 26 -5(32/9)#

#k = 74/9#

Use equation [1] to solve for #r^2#

#(9-32/9)^2+(8-74/9)^2=r^2#

#r^2 = 2405/81#

Because the area of a circle is:

#"Area" = pir^2#

We need only multiply by #pi#

#"Area" = 2405/81pi#