This is the third and final part of the how to think about co-ordinate geometry series. Due to a failure of calendar-reading, I appreciate this is going out a few days *after* the C1 and C2 exams, but hey-ho. If you're relying on this blog for your revision tips, you should have asked for me to bump it up the schedule.

Never mind that! On with circles.

When I ask my students what a circle is, I get a range of true replies, very few of which are useful. It's a round thing. Something with an infinite number of sides. Something with an infinite number of lines of symmetry (the only one of these that uniquely defines it, I think). But when I ask them how you draw a circle, they all know: you get a compass ('oh, that pointy thing, stab stab.') Here's the useful definition of a circle:

A circle is all of the points in a plane that are a fixed distance $r$ from a given centre $C$.

You stab the centre $C$ with your compass, put the pencil a distance $r$ away (the radius) and swoosh it around. Presto, a circle.

So, if the centre of a circle is at the point $(a,b)$, and any point on the edge of the circle is at $(x,y)$, you know that the distance between the two points is $r$. What's that tune I hear playing? It's got a sort of Greek style to it. That's right, it's the Pythagoras Theme! The distance between the two points is $\sqrt { \left( x - a \right)^2 + \left ( y - b \right)^2 }$, which we know is the same as $r$. Squaring both of them gives the equation of a circle:

$$\left( x - a \right)^2 + \left ( y - b \right)^2 = r^2$$

**You can generate the equation of a circle simply by applying Pythagoras**.

Can you complete the square? If not, you should buy my book, because you need to know how to complete the square to turn the expanded form of the equation of a circle $x^2 + y^2 - 2ax - 2by + a^2 + b^2 - r^2 = 0$ into the brackety form $(x-a)^2 + (y-b)^2 = r$.

Alternatively, if you know the template for the brackety form, you can simply multiply it out and match up the coefficients (make sure you have the same number of $x$s and $y$s and loose numbers in both equations.

Examiners have a mild obsession with tangents to a circle. It's easy enough to deal with them, though, as long as you know that the tangent is at right angles to the radius. You can find the gradient of a radius by working out $m = \frac{ y-b }{ x - a}$, the change in the $y$-direction over the change in the $x$-direction; the tangent is perpendracula to that, so you work out $m_\perp = \frac{-1}{m}$.

Then you've got a gradient and a point on a line. If only you had some way of turning that into an equation.

I can hardly believe you're asking this: this is something you've been doing since GCSE. To find out where two curves cross, you just find out where the equations of both things are true - which probably means simultaneous equations.

Now, I have a bee in my bonnet about circle theorems. I worked as a professional mathematician for 10 years, looking at the Sun. In that time, the only chance I got to use a circle theorem was to help a cartoonist friend find the centre of a circle he'd drawn with a yoghurt pot. My friends: the Sun is *round*. If there's one place you'd expect to be able to use a circle theorem, it's on the Sun. They're really not tremendously useful bits of real-life maths, in my opinion.

However, they can be somewhat useful in... answering questions on circles. The main one you might need is that if you have a diameter of a circle and draw a line from each end of it to another point on the circle, you get a right angle. And vice-versa: if you have a right-angled triangle, the three points are on a circle with a centre in the middle of the hypotenuse.

I think that's the only one that might come up. Do let me know if I've missed any!

Edited 2014-09-21 to fix a link.