# How the Mathematical Pirate finds the centre and radius of a circle

“Arr?” said the student, really not sure.

“No, no, $r$,” said the Mathematical Pirate. “The centre is at C – or $(a,\, b)$, if you prefer – and the radius is $r$.”

“Gotcha. So, if you’ve got something like $x^2 + y^2 + 8x - 12y + 3=0$, how do you find the centre? My teacher said something about completing the square.”

“Aye,” said the Mathematical Pirate. “Ye could do that, if you were a land-lubbing, rigour-centric piratophobe!”

“Arr!”

“Arr, indeed. ‘Tis not the *Pirate* way. You want to find the centre of the circle? You differentiate!”

“So, you need to rearrange to get $y$ on its own?” A doubtful tone.

“No, no, no, just ignore the $y$. $2x + 8 = 0$, so $x=-4$.”

“Are you sure that’s kosher?”

“It works, matey. You can do the same by ignoring the $x$.”

“So $2y - 12 = 0$ and $y=6$ – so the centre is at $(-4, 6)$? Is that it?”

“Well, that’s where the centre is, right enough. It’s not *it*, though, you still need the radius.”

“Oh.”

“Try putting the $x$ and $y$ you just found into the original equation.”

“But that’s the centre of the circle, it’s not on the…”

“Just do it.”

“Fine. $16 + 36 - 32 - 72 + 3= -49.$ That doesn’t seem much help.”

“Not until you realise the radius is $7$, no, it doesn’t.”

“That gives you $-r^2$?”

“If the Ninja was here, he’d make you prove it.”

“If the Ninja was here, we’d both be dead.”

“Fair point.”

“So the circle works out to have a centre at $(-4, 6)$ just by differentiating the $x$-bit and $y$-bit and setting them to zero, while the radius is $7$ because when you stick in the centre you get $-r^2$?”

“‘Tis a miracle,” said the Pirate. “But it be true. Arr.”