What is $b$ in this "conic general form" equation of a circle? $x^2+y^2+4x-4y-17=0$

Question

Take the equation of a circle from this khanacademy video as an example:

$$x^2+y^2+4x-4y-17=0$$

$$a=x^2$$ $$b= ? $$ $$c=y^2$$ $$d=4x$$ $$e=-4y$$ $$f=-17$$

What is b equal to, and why did we "jump" over that?

Answer 1

From the standard form $$ax^2+bxy+cy^2+dx+ey+f=0$$ you see which letters correspond to which coefficients, which isn't the same as the entire terms.

For example, $a$ is the coefficient of $x^2$, not $a=x^2$. Comparing with: $$x^2+y^2+4x-4y-17=0$$ you get $a=1$, ..., and $f=-17$.

Now notice that there is no term containing $xy$, so the term of the form $bxy$ has which coefficient? That makes $b=\ldots$.

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Hint:

$$x^2+y^2+\color{blue}{0xy}+4x-4y-17=0$$

Answer 2

$b$ is the coefficient of the $xy$ term. In this case it happens to be $0$. When $a=c$ and $b=0$, the conic is a circle.

Actually, it should be $ax^2+bxy+cx^2+dx+ey+f=0$, for the general conic,so that $a,b,c,d,e,f$ are the coefficients of the terms.

So we would have $a=1\\b=0\\c=1\\d=4\\e=-4\\f=-17$.