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The problem is: Let $A$, $B$ and $C$ be fixed points, and $α,β,γ$ and $κ$ are given constants, then the locus of a point $P$ that satisfies the equation $$α(AP)^2+β(BP)^2+γ(CP)^2=\kappa,$$ is a circunference. Prove it.

I need at least some hint to answer it, I tried using the distance between two points formula but I only get a mess of variables that show me nothing.

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    $\begingroup$ Writing $P=(x,y)$, if you were to expand everything (which you don't really have to do), you'd get an equation that's quadratic in $x$ and $y$, with no $xy$ term. Moreover, the coefficients on $x^2$ and $y^2$ would match. Therefore, ... $\endgroup$ – Blue Sep 16 '14 at 8:14
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Let $a=(a_x,a_y), b=(b_x,b_y), c=(c_x,c_y), P=(x,y)$
The equation is equivalent to $$\alpha(x-a_x)^2+\alpha (y-a_y)^2+\beta (x-b_x)^2+\beta (y-b_y)^2+\gamma(x-c_x)^2+\gamma(y-c_y)^2=\kappa$$ Rearrange to get, $$(\alpha+\beta+\gamma)x^2+(\alpha+\beta+\gamma)y^2-2(\alpha a_x+\beta b_x+\gamma c_x)x-2(\alpha a_y+\beta b_y+\gamma c_y)y=(\kappa-\alpha a_x^2-\alpha a_y^2-\beta b_x^2-\beta b_y^2-\gamma c_x^2-\gamma c_y^2)$$ To simplify the notation let $$ d_x=(\alpha a_x+\beta b_x+\gamma c_x)/(\alpha+\beta+\gamma)$$ $$ d_y=(\alpha a_y+\beta b_y+\gamma c_y)/(\alpha+\beta+\gamma)$$ $$ d=(\kappa-\alpha a_x^2-\alpha a_y^2-\beta b_x^2-\beta b_y^2-\gamma c_x^2-\gamma c_y^2)/(\alpha+\beta+\gamma)$$ divide the equation by $(\alpha+\beta+\gamma)$ and further rearrange the equation we'll get $$(x-d_x)^2+(y-d_y)^2=(d+d_x^2+d_y^2)$$

Therefore it's circunference center at $(d_x,d_y)$ with radius $\sqrt{d+d_x^2+d_y^2}$

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  • $\begingroup$ thank you so much, I was stuck on step 2(needed to simplify the notation). $\endgroup$ – mobzopi Sep 16 '14 at 9:21
  • $\begingroup$ hey could you explain me what you did after simplifying the notation? I tried it myself and I cannot get to the same result. Thanks! $\endgroup$ – mobzopi Sep 16 '14 at 9:46
  • $\begingroup$ Just complete square, after the division you get $x^2+y^2-2d_x x-2d_y y=d$ notice $(x-d_x)^2=x^2-2d_x x+d_x^2$ $\endgroup$ – gamma Sep 16 '14 at 9:52
  • $\begingroup$ Ok thanks, and when simpliflying the notation, why do yo divide both terms by /(α+β+γ)? $\endgroup$ – mobzopi Sep 16 '14 at 9:56
  • $\begingroup$ oh sorry, I'm blind, you used it to get rid of (α+β+γ) in the x^2 and y^2 terms right? $\endgroup$ – mobzopi Sep 16 '14 at 9:58

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