Depending on your AS-level maths exam board, you might encounter the equation of a circle in C1 (OCR) or C2 (everyone else). It's really just a restatement of Pythagoras' Theorem: saying $(x-a)^2 + (y-b)^2 = r^2$ is the same as saying "the square of the horizontal distance between $(a,b)$ and $(x,y)$ plus the square of the vertical distance is the same as the square of the direct distance" - which is, of course, the hypotenuse.

A typical question gives you an expanded form - maybe $x^2 + y^2 + 8x + 6y = 0$ - and asks you for the centre and the radius.

Of course, there's more than one way to do it.

This is the traditional method: you would say $x^2 + 8x \equiv (x + 4)^2 - 16$ and $y^2 + 6y = (x+3)^2 - 9$, so the left hand side works out to be $(x+4)^2 + (y+3)^2 - 25$ - meaning the circle has centre $(-4,-3)$ and radius $5$.

Which is all well and good, as long as you're comfortable completing the square. However, you can apply some different tools, too.

I'm going to skip the details of what's going on with the partial differentiation method and cut straight to the chase: to find the $x$-coordinate of the centre, ignore the $y$s and differentiate: $2x + 8 = 0$, so $x = -4$. Now there's a coincidence. Surely it can't work for $y$ as well?

Of course it works for $y$ as well. Ignore the $x$ and differentiate: $2y + 6 = 0$, so $y=-3$. Brilliant. Now, what about the radius?

Ah, that's the clever bit: you rearrange, if you need to, so that you have an expression that's equal to $0$, and put your values of $x$ and $y$ for the centre into that expression. Here, we don't need to rearrange (it already equals $0$), so we work out $(-4)^2 + (-3)^2 + 8(-4) + 6(-3)$ to get $16 + 9 -32 - 18 = -25$. Quick, look over there while I take the negative of it and then take a square root! The radius is 5.

Aw, you spotted it. Right: when you rearrange the circle equation, you get the expression $(x-a)^2 + (y-b)^2 - r^2$ as the left hand side. If you put $x=a$ and $y=b$ in, the brackets vanish and you're left with $-r^2$. Happy now?

Another option - very similar, in fact, to partial differentiation - is to notice that $\frac{dy}{dx}=0$ directly above and below the centre; similarly, $\frac{dx}{dy}=0$ to the left and right.

If you differentiate $x^2 + y^2 + 8x + 6y =0$ with respect to $x$, you get $2x + 2y \frac{dy}{dx} + 8 + 6 \frac{dy}{dx} = 0$; substituting in $\frac{dy}{dx}=0$ gives you $2x + 8 = 0$, giving the $x$-coordinate of the centre as $-4$

With respect to $y$, you get $2y + 6 = 0$ after you set the derivatives equal to $0$. Those two equations give you the centre of the circle.

Pick a point, any point.

Evaluate $\frac{dy}{dx}$ at that point, even if it's not on the circle.

Find the line perpendicular to that gradient through the point.

Repeat those three steps with a point that's not on the line you just found.

These lines cross at the centre of the circle. Why?

## Cav

Great post! I think I will explore the final method with my further maths class.