Question #7a456

1 Answer
seph
Oct 9, 2014

A circle is in the form

(x−h)2+(y−k)2=r2

where (h,k) is the center and r is the radius

Substituting known points in the circle, we have

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

Since they are all equal to r2, we can interchange the equations
[1] = [2]
(2−h)2+(8−k)2=(−4−h)2+(6−k)2
⇒(4−4h+h2)+(64−16k+k2)=(16−8h+h2)+(36−12k+k2)

[1] = [3]
(2−h)2+(8−k)2=(1−h)2+(−1−k)2
⇒(4−4h+h2)+(64−16k+k2)=(1−2h+h2)+(1+2k+k2)

Move all to the left side and simplify

[1] = [2]
⇒(4−4h)+(64−16k)−(16−8h)−(36−12k)=0
⇒16+8h−4k=0
⇒8+4h−2k=0

[1] = [3]
⇒(4−4h)+(64−16k)−(1−2h)−(1+2k)=0
⇒66−2h−18k=0

[1'] 8+4h−2k=0
[2'] 66−2h−18k=0

Eliminate one of the variables by scaling any of the equations
such that the variable will have the same coefficient (different sign is OK) as with the other equation

[1'] 8+4h−2k=0
[2'] 132−4h−18k=0

Add (or subtract if different sign)

140−20k=0
140=20k
k=7

[1'] 8+4h−2(7)=0
4h=6
h=32

the circle is centered at (32,7)

Related questions
Impact of this question
740 views around the world
You can reuse this answer
Creative Commons License