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$A(-3,1), B(0,-5), P(X,Y)$

If $|AP| = 2|BP|$ prove that $x$ and $y$ satisfy the equation:

\begin{aligned} \ x^2+y^2-2x+14y+30 =0 \end{aligned}

I get as far as determining the co-ordinates like so

\begin{aligned} \ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2} \end{aligned}


\begin{aligned} \ x^2+y^2+2y+40-6x =0 \end{aligned}

Which gives me $(3, -1)$, this won't satisfy, is my method correct?

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up vote 3 down vote accepted

If we start from the equation $$ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2}, $$ and square both sides, we obtain the equivalent equation $$x^2+6x+9+y^2-2y+1=4x^2+4y^2+40y+100,$$ which simplifies to $$3x^2+3y^2-6x+42y+90=0,$$ which is equivalent to $$x^2+y^2-2x+14y+30=0.$$

Comment: Draw the line segment that joins $A$ to $B$. It is fairly easy to see that $(-1,-3)$ is the point on this line segment that is twice as far from $A$ as it is from $B$. So $(-1,-3)$ ought to be on our circle, and indeed it is. That provides a partial check on the correctness of our computations.

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Shouldn't it be 2x^2+2y^2+20y+50? – Rollo Montgomery Konig-Brock Jan 22 '12 at 19:45
No, when you square both sides you need to square the "$2$". If $a=2b$ then $a^2=(2b)^2=4b^2$. – André Nicolas Jan 22 '12 at 19:50

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