# There's More Than One Way To Do It: Equations of a circle

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.

### Complete the square

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.

### Partial differentiation

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.

#### Wait a minute… I saw that!

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?

### Implicit differentiation

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.

### Implicit differentiation for smartarses

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?

## A selection of other posts

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