How can I "solve" or simplify quadratics of two variables?

Question

Given a multivariate polynomial equation like $$xy - bx - ay - ab = 0$$ how do I simplify it, make sense of it, or characterize it?

My attempts so far are below.

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Attempt 1

Viewing $x,y$ as variables and $a,b$ as constants, it's obviously of degree 2. Thus, I won't be able to find any linear solutions in $x$ or $y$. But I should be able to somehow make some sense of it, more than in the current form.

One approach I've tried is finding roots. As a quadratic, it should be expressable as a product of monomials. I've looked up on this site and elsewhere, and there seems to be no standard method to do that for multivariate polynomials, but we can treat $x,y$ individually (viewing the other as a constant), giving something like $$x = \frac {a(y-b)}{y-b} \\ = a$$ for $y \neq b$. But this same argument also gives $y = b$ (!), violating our condition.

A form like $$(x-p)(x-r)(y-q)(y-s) = 0$$ would be ideal. The above suggests it would simply be $$(x - a)(y - b) = 0$$ which seems to meet the conditions: $[x = a \land y \neq b] \lor [x \neq a \land y = b]$.

Multiplying out, we get $$(x - a)(y - b) = xy - bx - ay + ab$$ and so we've (in some ways) solved the problem.

Still remaining question: What is the locus of $x,y$ that meet that equation?

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The above seems to have found a simplification, but through a confused and roundabout path (and still incomplete). Is there a more direct or disciplined approach? Or a way to generalize this? When I see a polynomial in one variable, I immediately know what to do (for some definition of do). What about multivariate?

How can I "solve" these types of equations? I request that answers don't just solve this question, but show me the method (or point me to a good reference). Which courses is this normally covered in?

(Note: I've searched math.SE and other sites and found posts like this, this, this, and this none of which helped me.)

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Attempt 2

As pointed out in the comments, there's an error in the above. Rather, it should be: $$ x = a \frac {y+b}{y-b}, \quad y \neq b \\ y = b \frac {x+a}{x-a}, \quad x \neq a $$ so, if I simplified correctly, $$ x = a \left ( \frac {1 + \frac{x+a}{x-a}} {-1 + \frac{x+a}{x-a}} \right ) $$ which leaves me just as confused! What next?

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Update

Let's start a fresh:

$$xy−bx−ay−ab=0$$

Trying to isolate and solve for $x$ or $y$ is a mess, as the above shows. Lesson: We can't always solve for $x$! Instead, we can notice that:

1. This is a second degree polynomial in two variables
2. and therefore either a conic or degenerate conic
3. which are often best characterized not by $x=...$ but by $(\frac x p - h)^2...$ or something similar.

How do we do that? Let's try to factor our polynomial (or at least the higher degree terms of it) using an analog of completing the square: To get an $xy$ term, our solution will need to look like $$(x - p)(y - q) = r.$$ Then we must have $q =b$, because, when multiplied out, the only coefficient for $x^1$ is $q$. Likewise, $p = a$, giving $$(x-a)(y-b) = r.$$

Finally, since $$(x-a)(y-b) = xy -bx -ay + ab$$ then $r = 2ab$, giving $$(x-a)(y-b) = 2ab.$$ From knowledge of conics, this is a hyperbola with center $(a, b)$ and "scale factor" (I don't know the proper term) $2ab$. QED.