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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 the locus
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Write an equation as you read the problem
= .
Square both side
(x-3)^2 + (y-2)^2 = .( (x+1)^2 + (y-3)^2 )
Simplify it step by step
4 * ( (x-3)^2 + (y-2)^2 ) = (x+1)^2 + (y-3)^2
4*(x^2 - 6x + 9 + y^2 -4y + 4) = x^2 + 2x + 1 + y^2 - 6y + 9
4x^2 - 24x + 36 + 4y^2 - 16y + 16 = x^2 + 2x + 1 + y^2 - 6y + 9
The further simplification is routine bothering calculations, from which you will learn NOTHING.
It is a standard completing the square procedure.
So I will complete by referring to specialized online calculator
https://www.equationcalc.com/conics-section-calculator
Copy/paste the last equation to this calculator input port and get there the answer: final equation is
+ = .
It is the standard form equation of the circle with the center at the point (,) with the radius of .
Solved.
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It is really BAD problem to learn from.
My condolences . . .
Would somebody asked me about an educational value of this problem, I'd say it is zero.