I'm trying to demonstrate that, given a real number $k$, with $k>0$ and $k\neq 1$, the set $\{P \mid d(P,A) = k\cdot d(P,B)\}$ always represents a circle.
I simply gave the coordinates $A=(m,n)$, $B=(p,q)$ and $P=(x,y)$ and put into $d(P,A) = k \cdot d(P,B)$ and found the equation: $$(1-k^2)x^2 + (1-k^2)y^2 + (2k^2p-2m)x + (2k^2q-2n)y + (m^2+n^2-k^2p^2-k^2q^2)=0$$ that, written as $Ex^2 + Ey^2 + Fx + Gy + H = 0$ is a circle, with the respectives $E,F,G,H$.
I'm stuck at this point. The material shows that this is enough proof, but I'm not convinced because some questions before at the same list, I proved that $Ex^2 + Ey^2 + Fx + Gy + H = 0$ is a circle if, and only if, $E \neq 0$ and $F^2+G^2-4EH>0$.
As $k\neq1$, there's no problem with $E \neq 0$. But how to prove thaat $F^2+G^2-4EH>0$? The equations simply don't match... I tried some substitutions, but I'm ending with a lot of variables, with no relation to each other... I can't find a way around that.
Any insights?
Without the loss of generality you may assume that $A=(0,0)$ and $B=(b,0)$ (you can shift and rotate the axis without changing the geometry), then the equation you get is much simpler. $$x^2+y^2=k^2[(x-b)^2+y^2] \implies (1-k^2)x^2+(1-k^2)y^2+2bk^2x-b^2k^2=0.$$ Now you have $$E=1-k^2, F=2bk^2, G=0, H=-b^2k^2.$$ The condition $F^2+G^2-4EH>0$ can be easily verified.