A triangle has corners at (6,8), (5,4), and (3,2). What is the area of the triangle's circumscribed circle?

1 Answer
Douglas K.
Nov 15, 2016

A=85π2

Explanation:

The standard form for the equation of a circle is:

(x−h)2+(y−k)2=r2

where (x,y) is any given point on the circle, (h,k) is the center, and r is the radius.

Use the standard form and the three given points to write 3 equations:

(6−h)2+(8−k)2=r2 [1]
(5−h)2+(4−k)2=r2 [2]
(3−h)2+(2−k)2=r2 [3]

Set the left side of equation [1] equal to the left side of equation [2]:

(6−h)2+(8−k)2=(5−h)2+(4−k)2 [4]

Set the left side of equation [1] equal to the left side of equation [3]:

(6−h)2+(8−k)2=(3−h)2+(2−k)2 [5]

Use the pattern, (a−b)2=a2+3ab+b2 to expand the squares:

36−12h+h2+64−16k+k2=25−10h+h2+16−8k+k2 [6]
36−12h+h2+64−16k+k2=9−6h+h2+4−4k+k2 [7]

The k2andh2 terms cancel:

36−12h+64−16k=25−10h+16−8k [8]
36−12h+64−16k=9−6h+4−4k [9]

Collect the constant terms into a single term on the right:

−12h−16k=−10h−8k−59 [10]
−12h−16k=−6h−4k−87 [11]

Collect all of the h terms into a single term on the right:

−16k=2h−8k−59 [12]
−16k=6h−4k−87 [13]

Collect all of the k terms into a single term on the left:

−8k=2h−59 [14]
−12k=6h−87 [15]

Divide equation [14] by -8 and equation [15] by -12

k=−14h+598 [16]
k=−12h+8712 [17]

Set the right side of equation [16] equal to the right side of equation [17]:

−14h+598=−12h+8712 [18]

Solve for h:

14h=8712−598

h=873−592

h=−12

Substitute −12 for h into equation [17]:

k=14+8712

k=152

Substitute the values of h and k into either equation, [1], [2], or [3]. I will use equation [1]:

(6−−12)2+(8−152)2=r2

(132)2+(12)2=r2

r2=1704=852

The area of the circle is πr2:

A=85π2

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