TITLE: Why does dividing both sides of this system of equations to each other yields infinite "incorrect solutions"?
QUESTION [24 upvotes]: This might be basic, but I'm really bad at basic math. I'm trying to solve the following system of equations:
$$\sqrt{x^2+y^2}\cdot \left(x-5\right)=6x+y \tag{1},$$$$\\\sqrt{x^2+y^2}\cdot \left(y-1\right)=6y-x-2 \tag{2}$$
I put them in Wolfram Alpha to test the result, and it yields 3 solutions, which I assume is true. All nice and dandy.
But then I couldn't find out how to move forward. So I decide to divide $(2)$ by $(1)$. which gives:
$$\frac{y-1}{x-5}=\frac{6y-x-2}{6x+y} \tag{3}$$
After a few calculations, I figured that this is an equation for a circle. $$y^2+29y=-x^2+9x+10 \tag{4}$$ which implies that there's an infinite number of solutions. Which means that I'm wrong.
So can anybody tell me what I did wrong? And if possible teach me how to solve the system of equations please? Thank you! :D

REPLY [2 votes]: Contrary to your thinking, division is not the guilty. By the way, it doesn't take any division to combine the equations in
$$(6x+y)(y-1)=(6y-x-2)(x-5)$$
which is indeed that of a circle. (You multiply the first by $y-1$ and the second by $x-5$ and equate the LHS.)
But what you forgot is that you have a system of equations and dropping one of them enlarges the solution set. What you should consider is the augmented system
$$\begin{cases}\sqrt{x^2+y^2}\cdot \left(x-5\right)=6x+y ,\\\sqrt{x^2+y^2}\cdot \left(y-1\right)=6y-x-2,
\\x^2+y^2-9x+29y-10=0.\end{cases}$$

The two initial curves have three intersection points (two of them are close together, mind the small blue loop). And when you intersect either curve with the circle, you find four intersections. These extra points appear where $y=1$ or $x=5$, which makes one curve equation vanish.