A triangle has corners at #(7 ,3 )#, #(5 ,8 )#, and #(4 ,2 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-26T19:01:50

#color(blue)((5365pi)/578)# units squared.

The vertices of the triangle are all points on the circumference of the given circle. The equation for a circle is given by:

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

Where #bbh and bbk# are the #bbx# and #bby# coordinates of the centre respectively anf #bbr# is thr radius.

We can make 3 equations from this:

#(7-h)^2+(3-k)^2=r^2 \ \ \ \ \ [1]#

#(5-h)^2+(8-k)^2=r^2 \ \ \ \ \ [2]#

#(4-h)^2+(2-k)^2=r^2 \ \ \ \ \ [3]#

Subtracting #[3]# from #[2]#

#69-2h-12k=0 \ \ \ \ \[4]#

Subtract #[2]# from #[1]#

#10k-4h-31=0 \ \ \ \ \ [5]#

From #[5]#:

#k=(4h+31)/10#

Substituting in #[4]#

#69-2h-12((4h+31)/10)=0#

#69-2h-(6(4h+31))/5=0#

#345-10h-24h-186=0#

#h=159/34#

Substituting this in #[5]#

#10k-4(159/34)-31=0#

#k=(31+4(159/34))/10#

#k=169/34#

So we have the centre, we now find the radius:

Using vertex #(4,2)#

#(4-159/34)^2+(2-169/34)^2=r^2#

#r^2=5365/578#

Area:

#(5365pi)/578#

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