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. 

$$|AP|^2=(x+1)^2+(y-3)^2$$
$$|BP|^2=(x-2)^2+(y-7)^2$$
The equation of $C$ may be worked out as follows:
$$|AP|=\alpha|BP|$$
$$|AP|^2=\alpha^2|BP|^2$$
$$(x+1)^2+(y-3)^2=\alpha^2((x-2)^2+(y-7)^2)$$
$$x^2+2x+1+y^2-6y+9=\alpha^2(x^2-4x+4+y^2-14y+49)$$
$$(\alpha^2-1)x^2+(\alpha^2-1)y^2-(\alpha^2+\tfrac12)(4x)-(\alpha^2-\tfrac37)(14y)=10-53\alpha^2$$
$$x^2+y^2-\frac{2(\alpha^2+\frac12)}{\alpha^2-1}(2x)-\frac{7(\alpha^2-\frac37)}{\alpha^2-1}(2y)=\frac{10-53\alpha^2}{\alpha^2-1}$$
Let $M=\frac{2(\alpha^2+\frac12)}{\alpha^2-1}$, $N=\frac{7(\alpha^2-\frac37)}{\alpha^2-1}$ and $K=\frac{10-53\alpha^2}{\alpha^2-1}$, then
$$x^2+y^2-2Mx-2Ny=K$$
$$(x-M)^2+(y-N)^2=K+M^2+N^2$$
This last equation of $C$ is now in the canonical form for a circle.
