SOLUTION: Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P, PA^2 + PB^2 + PC^2 = 3PQ^2 + k. If A = (4,-1), B = (-

Algebra ->  Circles -> SOLUTION: Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P, PA^2 + PB^2 + PC^2 = 3PQ^2 + k. If A = (4,-1), B = (-      Log On


   

Question 1202365: Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P,
PA^2 + PB^2 + PC^2 = 3PQ^2 + k.
If A = (4,-1), B = (-3,1), and C = (5,-3), then find the constant k.

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let P be (x,y).  Then

PA^2 + PB^2 + PC^2 = 

= (x-4)^2 + (y+1)^2 + (x+3)^2 + (y-1)^2 + (x-5)^2 + (y+3)^2 =

= 3x^2 + 3y^2 - 12x + 6y + 61 = 

=3(x^2 + y^2 - 4x + 2y) + 61 =

= 3((x-2)^2 + (y+1)^2) + 46.


This shows that if  Q = (2,-1),  then  PQ^2 = (x-2)^2 + (y+1)^2.

And so  PA^2 + PB^2 + PC^2 = 3PQ^2 + 46.


ANSWER.  k = 46.

Solved.