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Difficulty: MediumLoci and Geometric Constructions

Two fixed points in a Cartesian plane are given as A(−1,2)A(-1, 2) and B(3,4)B(3, 4). A point P(x,y)P(x, y) moves in the plane such that PA2+PB2=26PA^2 + PB^2 = 26. Which of the following equations represents the locus of PP?

  1. x2+y2−2x−6y+2=0x^2 + y^2 - 2x - 6y + 2 = 0Answer
  2. B
    x2+y2+2x+6y+2=0x^2 + y^2 + 2x + 6y + 2 = 0
  3. C
    x2+y2−4x−12y+4=0x^2 + y^2 - 4x - 12y + 4 = 0
  4. D
    x2+y2−2x−6y+30=0x^2 + y^2 - 2x - 6y + 30 = 0

Answer

The equation representing the locus of PP is x2+y2−2x−6y+2=0x^2 + y^2 - 2x - 6y + 2 = 0.
Using the Cartesian coordinate distance formula, PA2=(x+1)2+(y−2)2PA^2 = (x+1)^2 + (y-2)^2 and PB2=(x−3)2+(y−4)2PB^2 = (x-3)^2 + (y-4)^2. Adding these together yields 2x2+2y2−4x−12y+302x^2 + 2y^2 - 4x - 12y + 30. Setting this equal to 26 gives 2x2+2y2−4x−12y+4=02x^2 + 2y^2 - 4x - 12y + 4 = 0. Dividing the entire equation by 2 yields x2+y2−2x−6y+2=0x^2 + y^2 - 2x - 6y + 2 = 0, which is the correct locus equation.

Step-by-Step Solution

1
Express PA2PA^2 using the distance formula between P(x,y)P(x, y) and A(−1,2)A(-1, 2).
PA2=(x−(−1))2+(y−2)2=(x+1)2+(y−2)2=x2+2x+1+y2−4y+4=x2+y2+2x−4y+5PA^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5
The square of the distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is (x2−x1)2+(y2−y1)2(x_2 - x_1)^2 + (y_2 - y_1)^2.
2
Express PB2PB^2 using the distance formula between P(x,y)P(x, y) and B(3,4)B(3, 4).
PB2=(x−3)2+(y−4)2=x2−6x+9+y2−8y+16=x2+y2−6x−8y+25PB^2 = (x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 = x^2 + y^2 - 6x - 8y + 25
Expanding the distance squared formula for point BB.
3
Sum PA2PA^2 and PB2PB^2 and equate to the given constant 26.
(x2+y2+2x−4y+5)+(x2+y2−6x−8y+25)=26  ⟹  2x2+2y2−4x−12y+30=26(x^2 + y^2 + 2x - 4y + 5) + (x^2 + y^2 - 6x - 8y + 25) = 26 \implies 2x^2 + 2y^2 - 4x - 12y + 30 = 26
Substitute the algebraic expressions into the locus condition PA2+PB2=26PA^2 + PB^2 = 26.
4
Rearrange into general circle form and divide through by 2.
2x2+2y2−4x−12y+4=0  ⟹  x2+y2−2x−6y+2=02x^2 + 2y^2 - 4x - 12y + 4 = 0 \implies x^2 + y^2 - 2x - 6y + 2 = 0
Subtract 26 from both sides and divide all terms by 2 to obtain the standard Cartesian equation.

Key Concept

Locus of a point with sum of squared distances to two fixed points equal to a constant
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Topics

Loci and Geometric Constructions

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