Question

Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, −1) is equal to 20

Answer 1

Let P(h, k) be the moving point

Let the given point be A(3, 5) and B(1, –1)

We are given PA² + PB² = 20

⇒ (h – 3)² + (k – 5)² + (h – 1)² + (k + 1)² = 20

⇒ h² – 6h + 9 + k² – 10k + 25 + h² – 2h + 1 + k² + 2k + 1 = 20

(i.e.) 2h² + 2k² – 8h – 8k + 36 – 20 = 0

2h² + 2k² – 8h – 8k + 16 = 0

(÷ by 2) h² + k² – 4h – 4k + 8 = 0

So the locus of P is x² + y² – 4x – 4y + 8 = 0