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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
Solution
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
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