A conjecture both deep and profound Needs a proof that the circle is round. In a paper by Erdős Written in Kurdish A counter-example is found.

One of my favourite questions to ask students is “what’s a circle?” because I get to play “so that means this is a circle!” until they give me the right definition. I had my jerkiness thrown back at me the other day, when my student told me “it’s a shape with an infinite number of edges and and infinite number of lines of symmetry,” which is a necessary and sufficient definition - but is entirely unhelpful in terms of figuring out what a circle is.

The clue I normally give after a few minutes is ‘how do you draw a circle?’ The way a pair of compasses works is by tracing out all of the points a fixed distance $r$ from a central point $C$. That’s the definition of a circle, and makes everything easier.

Pythagoras and the circle

Wait, I can hear you say, you’re trying to catch me out: Pythagoras is about triangles, not circles! Well spotted, gold star. All the same, Pythagoras is part of the circle formula because the definition of the circle relies on distance. The distance between a point $(x,y)$ on the edge of the circle from the centre $(a,b)$ is $\sqrt{ (x-a)^2 + (y-b)^2 }$, which we know is $r$. That’s a bit awkward, but we can rearrange it to be:

\[(x-a)^2 + (y-b)^2 = r^2\]

It’s Pythagoras’ Theorem, where $r$ is the hypotenuse and $(x-a)$ and $(y-b)$ are the legs. If you have an equation that looks like that, you can say straight off that the centre is at $(a,b)$ and the radius is $r$.

Sometimes you get given something like $x^2 + y^2 + 4x - 8y - 16 = 0$ and have to figure out the centre and radius. That’s just a case of completing the square - you say $x^2 + 4x = (x + 2)^2 - 4$ and $y^2 - 8y = (y-4)^2 - 16$, so it becomes:

$ (x + 2)^2 - 4 + (y-4)^2 - 16 - 16 = 0 $, which simplifies to $ (x+2)^2 + (y-4)^2 = 6^2$. The centre is at $(-2,4)$ and has radius 6.


For some reason, everyone I’ve taught this topic in the last couple of weeks has had the same brain-freeze over intersections, so I’m going to flag it up right here in bold:

To find out where two curves intersect, you solve their equations simultaneously.

A curve (in this context) is anything you can draw from an equation - a circle, a line, a parabola, a cubic, anything at all. So, if a question asks where a line intersects a circle, you just solve the simultaneous equations.

So, if you wanted to know where the circle from before crosses the line $y = 2x$, you’d just replace the y with 2x to get:

$ (x+2)^2 + (2x-4)^2 = 36$ $ (x^2 + 4x + 4) + (4x^2 - 16x + 16) = 36$ $ 5x^2 - 12x - 16 = 0$

That doesn’t factorise, so you use the quadratic formula (which you know, of course):

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{464}}{10}$, giving about 3.35 and -0.95.

Circle theorems

You also need to know a few circle theorems:

  • The tangent to a circle is perpendicular to the radius at that point, which means you can find the gradient of a tangent by working out the gradient of the radius and vice versa;
  • The angle opposite a diameter is a right angle, which again can give you two perpendicular lines (and a right-angled triangle)
  • If the tangents from two points (A and B) on a circle meet at a third point (P), the distances AP and BP are the same.

If you know and apply these ideas to any circles question, you should pick up most of the marks. Anything I missed? Let me know in the comments.