SOLUTION: Please help: A point P(x,y) moves in such a way that its distance from (3,2) is always one half of its distance from (-1,3). Find the equation of its locus.

Question 1071338: Please help: A point P(x,y) moves in such a way that its distance from (3,2) is always one half of its distance from (-1,3). Find the equation of its locus.
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We use the distance formula for two points
:
Let PA be the distance from P to A = (3,2) and
PB be the distance from P to B = (-1,3) then
:
PA^2 = (x-3)^2 + (y-2)^2
PB^2 = (x+1)^2 + (y-3)^2
:
Also PA^2 = PB^2 / 2 or 2 * PA^2 = PB^2
:
2 * [ (x-3)^2 + (y-2)^2 ] = (x+1)^2 + (y-3)^2
:
2 * ( x^2-6x+9 + y^2-4y+4 ) = x^2+2x+1 + y^2-6y+9
:
2x^2 -12x +18 +2y^2 -8y +8 = x^2 +2x +1 +y^2 -6y +9
:
x^2 -14x +y^2 -2y +16 = 0
:
complete the squares
:
1) (x-7)^2 + (y-1)^2 = -16 +50 = 34
:
equation 1 is a circle whose center is (7, 1) and radius is the square root(34) or 5.831
:

Answer by ikleyn(52803) (Show Source): You can put this solution on YOUR website!
.
The solution by "rothauserc" has an error in calculations.

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