This problem is generalized from problems 85 and 86 in section P.1 of "Calculus of a Single Variable" 8th edition by Larson, Hostetler, and Edwards.
We will find an equation describing all points \((x,y)\) such that (where \(d\) is the (Euclidean) distance function)
\[d((x,y),(x_1,y_1))=k\cdot d((x,y),(x_2,y_2))\]for given points \((x_1,y_1)\),\((x_2,y_2)\) and some positive constant \(k\). This describes all points whose distance from \((x_1,y_1)\) is \(k\) times its distance from \((x_2,y_2)\). This will turn out to be a line when \(k=1\) and a circle otherwise.
First we use the (Euclidean) distance formula and square each side.
\[(x-x_1)^2+(y-y_1)^2=k^2((x-x_2)^2+(y-y_2)^2)\]Next, expand everything and combine similar terms, using \(h=k^2\) to make the writing a little simpler.
\[x_1^2-hx_2^2+y_1^2-hy_2^2=(h-1)x^2+2(x_1-hx_2)x+(h-1)y^2+2(y_1-hy_2)y\]Then \(h=1\) so things simplify quite a bit. The \(x^2\) and \(y^2\) go away so what is left is a linear equation.
\[x_1^2-x_2^2+y_1^2-y_2^2=2(x_1-x_2)x+2(y_1-y_2)y\]Then we can factor.
\[2(x_1-x_2)x+2(y_1-y_2)y=(x_1+x_2)(x_1-x_2)+(y_1+y_2)(y_1-y_2)\]Notice that if \(x_1=x_2\) or \(y_1=y_2\), the equation simplifies to a single variable (the other) one equal to the average of the other coordinate. If both given points are the same, then we get \(0=0\) meaning every point satisfies the distance equation we started with. If both \(x_1\neq x_2\) and \(y_1\neq y_2\) then we can write it as
\[\frac{2x}{y_1-y_2}+\frac{2y}{x_1-x_2} =\frac{x_1+x_2}{y_1-y_2}+\frac{y_1+y_2}{x_1-x_2}\]This is where we divide by \(h-1\), which will make stuff messier.
\[\frac{1}{h-1}\left(x_1^2-hx_2^2+y_1^2-hy_2^2\right) =x^2+2\left(\frac{x_1-hx_2}{h-1}\right)x +y^2+2\left(\frac{y_1-hy_2}{h-1}\right)y\]On the right side, complete the square, and move the constant terms onto the left side.
\[\frac{1}{h-1}\left(x_1^2-hx_2^2+y_1^2-hy_2^2\right) +\left(\frac{x_1-hx_2}{h-1}\right)^2+\left(\frac{y_1-hy_2}{h-1}\right)^2 =\left(x+\frac{x_1-hx_2}{h-1}\right)^2+\left(y+\frac{y_1-hy_2}{h-1}\right)^2\]Everything on the left side is now a constant and the right side tells us that this is the equation of a circle. By using some more algebra, we can simplify the constant on the left and write the right side a little closer to how the standard circle equation looks.
\[\left(x-\frac{hx_2-x_1}{h-1}\right)^2+\left(y-\frac{hy_2-y_1}{h-1}\right)^2 =\frac{h}{(h-1)^2}\left[(x_2-x_1)^2+(y_2-y_1)^2\right]\]This shows that the radius depends on \(k\) and the distance between the 2 given points, which makes sense (if the points were moved but the distance apart stayed the same, intuitively it would make sense for the circle to just move without changing size). Lastly, put \(k^2\) back in.
\[\left(x-\frac{k^2x_2-x_1}{k^2-1}\right)^2 +\left(y-\frac{k^2y_2-y_1}{k^2-1}\right)^2 =\frac{k^2}{(k^2-1)^2}\left[(x_2-x_1)^2+(y_2-y_1)^2\right]\]