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Given point A(2,5) and point B(2,-1) find the set of points whose distance from A is double the distance from B. I attempted to solved it as follows: $$\left(x-2\right)^2+\left(y-5\right)^2=2\left(\left(x-2\right)^2+\left(y+1\right)^2\right)$$ This yields: $$\left(x-2\right)^2+\left(y+7\right)^2=72$$

However, the proposed solution is: $$\left(x-\frac{2}{3}\right)^2+\left(y+3\right)^2=\frac{112}{9}$$

I can't seem to figure out where I went wrong.

Daniel Oscar
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    Note that $(x-2)^2+(y-5)^2$ is not the distance from $A$, but the squared distance from $A$. So, it is $4$ instead of $2$ on the RHS. – SMM Sep 11 '18 at 14:34
  • I suppose you're right, for the circumference of $\left(x-2\right)^2+\left(y+3\right)^2=16$ seems to be correct: since both A and B lie on x=2, the distance from B to P(2, 1) -- which belongs to the circumference -- is 1/3 that of the A. However, that is not the textbook solution... Is the textbook solution wrong? – Daniel Oscar Sep 11 '18 at 15:59
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    Yes, the solution is wrong. The correct solution is $(x-2)^2+(y+3)^2=16$. – SMM Sep 11 '18 at 17:47

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