Deriving the equation of a circle

Question

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.

Answer 1

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 $$\frac{5\alpha}{1-\alpha^2}$$

Answer 2

$$|AP|^2=(x+1)^2+(y-3)^2$$ $$|BP|^2=(x-2)^2+(y-7)^2$$ 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.