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

2 Answers
Narad T.
Jun 7, 2017

The area of the circumscribed circle is =78.54

Explanation:

To calculate the area of the circle, we must calculate the radius r of the circle

Let the center of the circle be O=(a,b)

Then,

(4−a)2+(7−b)2=r2.......(1)

(3−a)2+(4−b)2=r2..........(2)

(8−a)2+(9−b)2=r2.........(3)

We have 3 equations with 3 unknowns

From (1) and (2), we get

16−8a+a2+49−14b+b2=9−6a+a2+16−8b+b2

2a+6b=40

a+3b=20.............(4)

From (2) and (3), we get

9−6a+a2+16−8b+b2=64−16a+a2+81−18b+b2

10a+10b=120

a+b=12..............(5)

From equations (4) and (5), we get

12−b+3b=20

2b=8

b=82=4

a=12−b=12−4=8

The center of the circle is =(8,4)

r2=(4−a)2+(7−b)2=(4−8)2+(7−4)2

=16+9

=25

The area of the circle is

A=π⋅r2=25⋅π=78.54

Dean R.
Jun 29, 2018

(4−3)2+(7−4)2=10

(8−3)2+(9−4)2=50

(8−4)2+(9−7)2=20

πr2=π(10)(50)(20)4(10)(50)−(20−10−50)2=25π

Explanation:

Here's the shortcut.

The circumcircle is just the circle through the three vertices; the triangle almost doesn't matter. Except, miraculously, the circumradius r equals the product of the triangle sides a,b,c divided by four times the triangle's area A.

r=abc4A

It's much more useful squared, and we're looking for πr2 anyway.

πr2=πa2b2c216A2

The coordinates give the squared distances easily. Archimedes' Theorem relates the squared distances to the triangle area:

16A2=4a2b2−(c2−a2−b2)2

So,

πr2=πa2b2c24a2b2−(c2−a2−b2)2

We form the squared distances from pairs of points (4,7),(3,4),(8,9)

a2=(4−3)2+(7−4)2=10

b2=(8−3)2+(9−4)2=50

c2=(8−4)2+(9−7)2=20

πr2=π(10)(50)(20)4(10)(50)−(20−10−50)2=25π

Related questions
See all questions in Area of Inscribed Triangle
Impact of this question
1754 views around the world
You can reuse this answer
Creative Commons License