What exactly are circle theorems then? Well, they are properties that show relationships between certain angles in the geometry of a circle. We use these theorems alongside prior knowledge of other angle properties to calculate any angles that might be missing, without the need for a protractor.

There are seven main circle theorems:

Alternate segment circle theorem

The angle at the centre circle theorem

Angles in the same segment circle theorem

Angle in a semi-circle theorem

Chord circle theorem

Tangent circle theorem

Cyclic quadrilateral circle theorem

In this session, we are going to be looking at circle theorems and looking at the different types of circle theorems available. We're going to be working through each one bit by bit and then hopefully by the end of the session session, you'll be able to start using these in more context exam style questions. We won't be applying these to exam questions in this session. Essentially, we are going to be learning the different types of circle theorems first. So what I would suggest you do before you get started on this section is to go back to the different types of the parts of a circle.

So arc, sector segments, chord and all those types of areas of a circle, because it's really important that you understand those before you move on to these. So, circle theorem one.

So a tangent to a circle makes a right handle with a radius at that point. So basically what they're saying is if you have a circle and you've got a radius and you've got a tangent, the angle between the radius and the tangent makes 90 degrees. So, again, a radius, remember it's from the centre to the circumference and a tangent SKIMS past the circle at one point only, so that angle there is 90 degrees. Now, with a lot of these circle zone questions, what you will find is never really very rarely asked to just look at one circle zone. They could implement or make you go to your question where you are using quite a few of these theorems in one go.

So this might be the first step of working out. But they do like to ask you for your reasons for each stage. So you do need to remember these. And my suggestion is potentially have a diagram on one side of a flashcard and have the actual reason on the other. And I've tried to make these reasons as easy as possible because there are some sort of mathematic books and things like that that overcomplicate the reasoning.

You need a basic reason with your circle zone. So this is the first one. The angle between the radius and the tangent is 90 degrees.

We look at circle theory too. So we've got a circle here and these green lines are the radius, or each individual one is called the radius. And if you remember parts of the circle, the radius is basically always equal because it's coming from the centre to the circumference, which is the same length throughout your circle. So if we look at this first triangle here that I've made, that's the chord, because we've got a line going from one part of the circumference to the other and it's not running through the centre. So we've got a chord and we've got our radi.

And this makes an isosceles triangle. One of the properties of an isosceles triangle is that these two sides will be equal, therefore the base angles are equal. And sometimes you can get these questions that again require you to work as part of your answer on these types of questions. So you would need to realise that if this is 60 degrees, or let's make it 55, this would also be 55 degrees. Okay.

Likewise are slightly more spread out isosceles. But again, you can see these two lengths are the same. Therefore these two angles at the bottom, the base angles, are equal. So again, a triangle formed by two radius is isosceles.

I have seen some students go into more complex reasoning, which is fine, which is to say an isosceles are formed between two radius and a chord. That's fine. But again, try and make your reasoning as simple as possible so that you don't panic in an exam when you're having to reason.

Okay? Circle theory. So if you have a triangle within a semicircle, it essentially makes a 90 degree angle. So you'll see here, that the diameter part. It is the hypotenuse of your triangle.

And so if you can form a triangle there, then this angle here at the circumference is 90 degrees. Likewise, if we look this way, we've got a 90 degree angle up here. So I always see it as a triangle tucked into a semicircle. But just remember that the hypotenuse has to be the diameter circle there and four. So this one is a slightly more different one.

So we've got a chord here. So, again, remember, a chord is a line that goes from one end of a circumference to the other and bisecting. Now, bisecting basically means cutting in half. So bisecting a chord is splitting the cord exactly through the centre. Okay?

So this is your diameter here and this is your chord. Okay. The angle between your cord and your diameter is 90 degrees. And this line will always have to be the diameter for it to be cutting it exactly to the centre. Okay?

Because halfway between the cord will be lying on that diameter. So an easier way to say that is a diameter bisect the cord at 90 degrees or at right angles. So this side would also be 90 degrees here. Those two are clients here. These little marks, they're just showing that these two sides are equal in length because you've cut it in half.

Okay, circle theorem five. So a lot of my students look at this as a rocket style shape, which is fine, because it does, on the whole tend to remind them of the circle theorem. But just please bear in mind that not all of these types of questions for circle theorem five end up looking like a rocket. And if I could drag this top corner and I move it, rotate it around circumference and make it lower and lower, it's still this circle there. It just doesn't look like a rocket anymore.

So just be very careful. But on the whole, this rocket does help students when it looks like the sort of shape. And again, I've seen this in combination with a different circle theorem. I think it might be circle of seven. And that's where you really do need to remember this theorem.

And this theorem basically states that the angle at the centre is double the angle at the circumference. So if A was 60 degrees, the angle at the centre would be 120 deg. Likewise, if this was 80 degrees here, then the angle at A would be half of that, which is 40 deg. So you are technically looking for some form of a rocket shape. Just be aware that sometimes that angle up here can be quite low down here.

So the oval shapes bit more squashed and less rocket like.

Circle six. So again, another way that my students have seen this is as butterfly wings. If you can see, these two triangles sort of form a butterfly for the angles in the same segment, these top two are the same as these bottom two. Okay? So angles in the same segment are essentially equal.

If B is 40 degrees here, it's going to be 40 degrees up here as well, but they have to be in the same segment.

Circle theorem seven. So here we have a four sided shape completely encased within the circle. All four corners are touching that circle. And so what's really important is understanding that this is a cyclic quadrilateral. The four sided shape within a circle and the opposite angles in a cyclic quadrilateral add up to 180.

So A plus C together makes 180 and D plus B together makes 180. So if A was 100 degrees, c will be 80 degrees and so on. Again, I've seen a rocket shape within this one. So you might have another sort of two lines here going in like that. You've got a rocket on the top.

So we would then say that those opposite angles are adding up to 180. Circle theorem eight. There's two more circle theorem eight is two tangents that are coming from a single point that touch your circle. Those two tangents are basically equal. And again, I've seen a question that combines the angle between a tangent and a centre that makes 90 degrees.

I've seen that combination as well. And sometimes you can be asked to find this top angle, sometimes you can be asked to find the centre angle. So these two tangents are the same length. And then the final circle theorem is the angle between a tangent and a chord. This is a more complex one, is equal to the angle subtended from the ends of the chord in the alternative segment.

In other words, if you have a tangent and then you have a triangle within your circle, the angle between the left hand side between the tangent and the side of the triangle, the chord is equal to the interior opposite angle in the triangle. So B is equal to the interior opposite angle. So it would be this one. Always work with the opposite. That's the most complex circle theorem that you need to memorise.

And that's awesome.