<s>[INST] Deriving the equation of a circle I have two points, $A=(-1,3)$ and $B=(2,7)$. There is a third point $P=(x,y)$.
I have found $|AP|$ and $|BP|$ in terms of $x$ and $y$. These are:
$|AP| = \sqrt{(-1-x)^2 +(3-y)^2}$
$|BP| = \sqrt{(2-x)^2 +(7-y)^2}$
I have shown that the set $S = \{P:|AP| = |BP|\}$ is a straight line and determined the slope and y-intercept by expanding out $|AP| = |BP|$ and simplifying to get:
$y= \frac{-3}{4}x+\frac{43}{8}$.
What I can't derive is an equation for the circle $C= \{P: |AP| = \alpha|BP|\}$. I know it should be of the form $(x-a)^2 + (y-b)^2 = \alpha^2$ but when I try to tackle this algebraically I get stuck. 
 [/INST] Change the coordinate system to map $A$ to the origin in order to facilitate algebraic manipulations.
$$\begin{align}
(\;\ x,\;y)&\rightarrow \;\;(x+1,y-3)\\
A(-1,3)&\rightarrow A'(\;\;\;\;\;0,\;\;0)\\
B(\;\;2,7)&\rightarrow B'(\;\;\;\;\;3,\;\;4)\\
P(\;\;x,y)&\rightarrow P'(x+1, y-3)=P(x', y')
\\\\
|A'P'|&=\alpha |B'P'|\\
A'P'^2&=\alpha^2 B'P'^2\\
x'\ ^2+y'\ ^2&=\alpha^2\left[(x'-3)^2+(y'-4)^2\right]\\
(1-\alpha^2)(x'\ ^2+y'\ ^2)+6\alpha^2x'+8\alpha^2y'-25\alpha^2&=0\\
x'\ ^2+y'\ ^2+\frac{6\alpha^2x'+8\alpha^2y'-25\alpha^2}{1-\alpha^2}&=0\\
\left(x'+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y'+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\frac{25\alpha^2}{1-\alpha^2}+\frac{25\alpha^4}{(1-\alpha^2)^2}\\
\left(x'+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y'+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\left(\frac{5\alpha}{1-\alpha^2}\right)^2\\
\text{Converting back to original coordinates,}\qquad\qquad\qquad\qquad&&\\
\left(x+1+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y-3+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\left(\frac{5\alpha}{1-\alpha^2}\right)^2\\
\color{red}{\left(x-\left(-\frac {1+2\alpha^2}{1-\alpha^2}\right)\right)^2+\left(y-\frac{3-7\alpha^2}{1-\alpha^2}\right)^2}
&\color{red}{=\left(\frac{5\alpha}{1-\alpha^2}\right)^2}\\
\end{align}$$
which is the equation of the circle $C$, the locus of $P$. 
Circle $C$ has its centre at
$$\left(-\frac {1+2\alpha^2}{1-\alpha^2},\frac{3-7\alpha^2}{1-\alpha^2}\right)$$
and radius of 
$$\frac{5\alpha}{1-\alpha^2}$$
</s>