"Let me see that!" commanded the Mathematical Ninja, looking at one of the Mathematical Pirate's blog posts. "That's... but that's..."
"It's not wrong!" said the Mathematical Pirate, smugly. "It just works!"
"But you're presenting it as magic, not as maths."
The Mathematical Pirate nodded eagerly. "Lovely magic! How does it work, then, clever-clogs?"
The Mathematical Ninja sighed. "It's not as complicated as all that. You know the circle $C$ is the locus of all of the points a fixed distance $r$ from a given centre $(a, b)$?"
"Terribly sorry, I got bored and stopped listening." He twitched his cutlass to remind the Mathematical Ninja not to try anything.
"In any case, the points where the curve has zero gradient are $(a, b+r)$ and $(a, b-r)$."
"You mean it goes flat at the top and the bottom."
Scowl. "Which means, if we differentiate the equation of the circle implicitly..."
"The equation of a circle is $x^2 + y^2 + px + qy + k = 0$" sing-songed the Pirate.
"An equation. But yes, differentiated with respect to $x$, you get $2x + 2y \dydx + p + q \dydx = 0$."
"And since $\dydx = 0$, that gives $2x + p = 0$ where the curve is flat just like I said."
The Mathematical Ninja ignored him. "You can do the same thing with $\diff xy$ to find where the curve is vertical."
The Mathematical Pirate picked at his fingernails. "Nobody likes a smart-arse, you know. Anyway, what about that radius? Why is that what is is, huh?"
"That, my sea-faring friend, is simple: your equation of the circle tells you where a point is a fixed distance from the centre. Pythagoras says that's $(x-a)^2 + (y-b)^2 - r^2=0$, which is equivalent to your, vastly inferior, form."
"How can it be inferior if they're equivalent, hm?" The cutlass rattled again.
"Anyway, sticking $x=a$ and $y=b$ into the left-hand side of this gives you $-r^2$, as required."
"I was once caught in the doldrums for 40 days and 40 nights with only a Daily Mail columnist for company. And even then I wasn't as bored as I am now. I'm going to go and sing a shanty now."