# Geometric locus

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.

• 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, ... – Blue Sep 16 '14 at 8:14

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 $\sqrt{d+d_x^2+d_y^2}$
• Just complete square, after the division you get $x^2+y^2-2d_x x-2d_y y=d$ notice $(x-d_x)^2=x^2-2d_x x+d_x^2$ – gamma Sep 16 '14 at 9:52