Question 1191709
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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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<pre>
Write an equation as you read the problem

    {{{sqrt((x-3)^2+(y-2)^2)}}} = {{{(1/2)*sqrt((x+1)^2 + (y-3)^2)}}}.


Square both side

    (x-3)^2 + (y-2)^2 = {{{(1/4)}}}.( (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 


    {{{(x-(13/3))^2}}} + {{{(y-(5/3))^2}}} = {{{(2/3)^2*17}}}.


It is the standard form equation of the circle with the center at the point ({{{13/3}}},{{{5/3}}})  with the radius of  {{{(2/3)*sqrt(17)}}}.
</pre>

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.