Lesson video

Hello there and thank you for choosing today's lesson.

My name is Dr.

Rowlandson and I'll be guiding you through it.

Welcome to today's lesson from the unit of Circle theorems. This lesson is called "The tangent at any point on a circle is perpendicular to the radius at that point." And that is a circle theorem, which by the end of today's lesson, we will be able to derive and use.

Here are some previous keywords that will be useful during today's lesson.

So you may want to pause the video if you want to remind yourself what these words mean and then press play when you're ready to continue.

The lesson is broken into two learning cycles.

In the first learning cycle, we're going to investigate tangents on a circular geoboard.

Then in the second learning cycle, we're going to introduce the circle theorem more formally.

Let's start off with investigating tangents with a circular geoboard.

A tangent to a circle is a line that touches the circle exactly once on the circumference of that circle.

For example, here we have a circular geoboard and a tangent at point P.

That line only touches the circle exactly once.

Here's a non-example.

We have the same circular geoboard.

We have a line that goes through point P, but because it goes through another point on the circle, it means it's not a tangent.

It should only touch the circle exactly once.

Here's another non-example.

This line segment does only touch the circle once, but if we extend it into a line, it would touch the circle twice, so it's not a tangent.

It needs to touch the circle exactly once.

Let's check what we've learned.

Which of these lines are tangents to their circle? Pause the video while you choose and press play when you're ready for answers.

The answers are B and C.

These are the only lines that touch their circle exactly once.

And D is a line segment, and if you extend that into a line, it would touch the circle twice.

Here we have a circular geoboard with a chord on it.

We can create chords that are parallel to each other on a circular geoboard, and we can do that by considering the positions of the pegs.

A parallel chord can be made to this one by using two pegs that are equidistant from the original chord, with both pegs either being above or below the original chord.

For example, if we go one peg up from each of the two points on the chord we can see there, we can create a new chord like this.

And that one is parallel to the original one.

We could do the same but going below and create this chord which is parallel to the other two.

And we can do it again to create another parallel chord.

All four of these are parallel.

But what if there's only one point remaining? Hmm.

Well, by adding two more pegs, we can make another parallel line segment, but is it another parallel chord? No, it's not.

That line segment is not inside the circle.

It doesn't go from one point in the circumference to another, so it's not a chord.

What we have made is a tangent, and this tangent is parallel to all the chords we've made inside the geoboard.

If you'd like to investigate this yourself and have access to this slide, there is a link at the bottom of this slide that takes you to a GeoGebra version of a circular geoboard with a chord and a tangent on there.

It looks something a little bit like this.

So let's check what we've learned.

Which diagrams show a chord that is parallel to the tangent of its circle? Pause while you choose and press play when you're ready for answers.

A and B.

So here we have a circular geoboard with 10 pegs equally spaced around the circle and we've made a tangent by adding some extra pegs to the circular geoboard.

We can use this to calculate angles related to polygons that go outside of the circle, such as this triangle here.

We can find the interior angles of this triangle by drawing this chord here to create a smaller triangle that is inside the circular geoboard and is similar to the original triangle.

Here is the original triangle which goes outside of the circular geoboard, and here's a smaller one which is inscribed in the circular geoboard.

Those two triangles have the same angles because they are similar.

So if we can find the interior angle of this smaller triangle, with all the vertices on the geoboard, then we can also find the angles of the larger triangle.

We can do that by drawing on line segments, OA and OC.

These are both radii of the circle.

And then we can try and find the angle AOC in the centre.

To do that, we can consider this sector here.

There are 10 points that are equally spaced around this circle, which means this sector is one-tenth of the circle.

Therefore, the angle at the centre here would be 36 degrees.

That's one-tenth of 360.

If we have four of these, then the angle at the centre would be 144 degrees.

So we have that angle.

We can use that angle and another circle theorem to work out another angle.

Using the circle theorem, "the angle at the centre of the circle is twice the angle at any point on the circumference," we can work out the size of angle ABC.

That'll be 72 degrees.

It's half of 144.

Now we know that angle, we can work out other angles.

For instance, AB is equal to BC, since points A and C are both equal distance from point B.

Therefore, the triangle ABC is an isosceles triangle.

So we could work out angles by doing 180 degrees subtract the 72 and divide by two, and that'll give us angle BAC and angle BCA as both being 54 degrees.

Now we have all the angles in the smaller triangle.

The angles in the bigger triangle are the same.

Since these two are similar, their angles are invariant.

So that means the other two angles, BED and BDE, are both 54 degrees as well.

Let's check what we've learned.

Here we have a 12-point circular geoboard.

Calculate the size of angle BOD.

Pause while you do that and press play for an answer.

Well, we can do this by first dividing 360 degrees by the number of pegs, 12.

That'll give 30 degrees.

And then we need to multiply that by four because the angle BOD takes up four of those equal-size sectors.

That gives 120 degrees.

So now could you please calculate the size of angle BCD? Pause while you do it and press play for an answer.

Angle BCD is equal to 60 degrees.

You can get that by using the circle theorem, "the angle at the centre of the circle is twice the angle at any point on the circumference." So now we have those two angles, which of these statements are correct? Pause the video while you choose and press play when you're ready for answers.

Statements C and E are correct.

The length of BC is equal to the length of CD, and also the length of BO is equal to the length of OD.

So with that in mind, calculate the size of the angle CDB.

Pause the video while you do that and press play when you're ready for answers.

The answer is 60 degrees.

You can get that by subtracting 60 degrees from 180 and dividing by two, and you get 60 degrees.

So could you please calculate the size of angle CAE? Pause the video while you do that and press play when you're ready for answers.

The answer is 60 degrees.

Those two triangles are similar to each other.

They're both equilateral triangles.

Here we have an eight-point circular geoboard with a radius marked on, and here we now have a diameter which is perpendicular to the radius.

We can consider other line segments that are parallel to this diameter.

For example, this chord here is parallel to the diameter.

So is this chord here.

And so is this line segment here, which is a tangent to the circle.

Now we have a pair of parallel lines and a transversal between them, we can see that using alternate angles on a pair of parallel lines, the angle between the radius and a tangent is also a right-angle.

So let's check what we've learned.

Here is a radius and a diameter of this 20-point circular geoboard.

Which of these statements explains how you know the radius and diameter are perpendicular to each other? Pause by choosing A, B, and C, and then press play when you're ready for answer.

The answer is B.

There are an equal number of points on the circumference between A and B as there are between B and C.

Let's now add some more elements to this diagram.

We have a line segment that goes from O to D, which is a point outside the circle, and we have a tangent which is parallel to the diameter.

Calculate the size of angle BDO.

Pause the video while you do this and press play when you're ready for answers.

The answer is 54 degrees.

The way we get to that is by first looking at angle OBD, which is 90 degrees because it is alternate between parallel lines with the other 90-degree angle that was given to us.

And then we can look at angle DOB.

That is 36 degrees because that angle is at the centre of a sector which is two-twentieths of the whole circle, and two-twentieths of 360 degrees is 36 degrees.

Therefore, the angle BDO would be 180 degrees subtract 90 subtract 36, which gives 54 degrees.

This time, we have a 10-point circular geoboard with a diameter marked on it from A to B.

Andeep wants to draw on a radius that is perpendicular to this diameter.

Its point at the circumference must be on a point of the geoboard.

Explain why Andeep cannot do this.

Pause while you write something down and press play when you're ready for an answer.

Here's a possible answer.

There is an even number of points on the geoboard between A and B.

This means that a radius perpendicular to AB must intersect in between two of these points.

Okay, it's over to you now for Task A! This task contains four questions and here is question one.

Pause while you do it and press play when you're ready for more questions.

And here is question two.

Pause while you do this and press play for the next question.

Here is question three.

Pause while you do this and press play when you're ready for question four.

And here is question four.

Pause while you do this and press play when you're ready for answers.

Okay, let's go through some answers.

In question one, you had to work out some missing angles.

They're all now on the screen.

Please pause the video while you check these against your own and press play to continue.

Then question two, you had an eight-point circular geoboard.

And in part A, you had to draw on a radius that was perpendicular to the diameter AB.

Well, you can see a radius in the diagram here and that is one possible answer.

You can also have the radius that is on the other side of that diameter, going to the point which is directly across from it.

For part B, you had to draw on two tangents that are both parallel to diameter AB, and you can see those two tangents on the diagram here.

And what you should notice is that the radius and tangent intersect and they are perpendicular to each other.

And we know that they are perpendicular to each other because the tangent is parallel to the diameter.

And alternate angles between parallel lines are equal.

So if it's a right-angle between the radius and diameter, it's also a right-angle between the radius and the tangent.

And question three.

You had a 12-point circular geoboard with some angles to work out.

Angle BAC is 90 degrees, and your justification is that the angle in a semicircle is a right-angle.

That's one of the circle theorems. Angle CBA is equal to 15 degrees, and we can justify that by saying interior angles in a triangle sum to 180 degrees.

Therefore, if we subtract 90 and 75 from 180, we get 15 degrees.

Angle ADF is also equal to 15 degrees.

Our justification there is by using the parallel lines.

Corresponding angles across a pair of parallel lines are equal.

Therefore, angle ADF is equal to angle ABC.

Then question four.

You had to calculate the size of angle AFO, justifying your answer.

Well, the answer is 30 degrees, and here is one way to get to the answer with some justifications.

Pause the video while you check this against your own and press play when you're ready to continue with the lesson.

Great work so far.

Now let's move on to the next part of this lesson where we're going to introduce the circle theorem more formally.

Here we have a circular geoboard and Alex thinks he's spotted something.

He says, "If you have a diameter, and a radius which is perpendicular to it and then a tangent which is parallel to the diameter, the tangent always intersects that radius on the circumference at 90 degrees." So we have a tangent and the radius which intersect at a right-angle.

Alex says, "Do all tangents and radii that meet at the circumference of a circle form a right-angle?" Well, yes, they do, and that's the circle theorem of today's lesson.

The circle theorem says that the tangent at any point on a circle is perpendicular to the radius at that particular point.

So let's check what we've learned.

Here we've got a diagram with a circle, a radius, a tangent, and another line segment.

We can see that one of the angles is 32 degrees.

Could you please find the size of the three angles indicated on the screen? Pause the video while you do it and press play when you're ready for answers.

Let's take a look at some answers.

Angle OBA is 90 degrees because a radius meets a tangent at 90 degrees.

The size of angle OBC is also 90 degrees because a radius meets a tangent at 90 degrees.

It's just the other side of that radius.

Could also reason that two angles that meet on a straight line also sum to 180 degrees, so if one's 90, the other one must be 90 as well.

Angle BOA is 58 degrees, which you can get by using angles in a triangle sum to 180 degrees.

So Alex says, "Is there a way of explaining why a tangent and a radius that intersect are perpendicular to each other?" We can use other circle theorems to help explain why an intersecting tangent and radius are always perpendicular to each other.

We can start with any chord to the circle.

Now this chord intersects the circumference twice and we've called them points A and B.

A line that extends in the chord will be helpful later.

Another circle theorem states that the perpendicular from the centre of a circle to a chord bisects the chord.

So if this radius here is perpendicular to the chord, it breaks into two equal lengths, like we can see here.

This perpendicular radius intersects the circumference of the circle at another point.

Let's call it point T.

Points A and B are both equidistant from point T.

Hmm.

We can then consider a second chord that is parallel to this first chord.

Points A and B will remain equidistant from T.

The perpendicular bisector of this new chord is the same as the first.

You can do it like this.

And then we can repeat this process again, but with points AB both at the single point T.

Like this.

So the circle theorem that says, "the tangent at any point on a circle is perpendicular to radius at that point" is just a special case of the circle theorem that says, "the perpendicular from the centre of a circle to a chord bisects the chord." And if you have access to this slide, you can click on the link on the bottom of the screen that takes you to a GeoGebra file which demonstrates this nicely.

It looks something a bit like this.

So let's check what we've learned.

On this diagram, CD is a tangent to the circle at point T, and OT is a radius of the circle.

So which of the statements are true? Pause while you choose and press play when you're ready for an answer.

The answer is B.

Angle CTO must be 90 degrees because the tangent meets the radius at 90 degrees.

Here's another question.

AB and CD are parallel to each other.

OT is a radius of the circle.

Point X is the intersection of AB and OT.

So based on that information, which of the statements from A to E are true? Pause while you choose and press play for an answer.

Here are the statements which are true.

Statement A is true because we can see the chord is bisected by the line segment from the centre.

We can see that by the hash marks on it.

If it's bisected, it means the line segment from the centre must be perpendicular to the chord.

Therefore, AXO must be 90 degrees.

If AXO is 90 degrees, so must also be CTO because the chord is parallel to the line segment CD.

And if we can see that it's 90 degrees and it meets the radius at the circumference, it means CD must be a tangent to the circle, and CD is perpendicular to OT.

Here's another question.

OT is a radius of the circle.

Which of the statements are true? Pause while you choose and press play when you're ready for an answer.

The answer is C, a degrees must not be equal to 90 degrees.

We can see that VW intersects the circle twice, which means VW is not a tangent.

And because VW is not a tangent, radius OT is not perpendicular to VW.

So now we know this circle theorem, we can use it to solve problems. In this diagram, OT, OC, and OD are all radii.

TD is a diameter and AB intersects the circle at T.

Let's work out the value of x.

Now to work out the value x, we may need to work out some other angles along the way.

Perhaps pause the video and think about what angles you could work out before we find the value of x, and then press play when you're ready to do this together.

Let's look at this together now.

AB is a tangent to the circle, and a tangent meets a radius at 90 degrees, which means ATO is 90 degrees.

This is because AB is a tangent, and the tangent at any point on a circle is perpendicular to the radius at that point.

Now if we know that all that angle is 90 degrees and some of that angle is 71 degrees, then angle CTO must be 19 degrees because we can do 90 subtract 71 to get 19.

Now let's look at the triangle that contains that 19-degree angle.

Two of its sides, OT and OC, are radii of the circle, which means they are the same length.

That triangle is isosceles, which means angle TCO is also 19 degrees.

Because OC and OT are equal in length, therefore the triangle is isosceles.

And then we have the information that TD is a diameter.

A diameter splits a circle into two semicircles, which means angle TCD must be 90 degrees.

And this is because TD is a diameter, and the angle in a semicircle is a right-angle.

So with that right-angle, we can see that it's made up of 19 degrees plus x degrees.

We can now work out the value of x by doing 90 subtract 19 to get 71 degrees.

So let's check what we've learned.

Based on this diagram, which of these statements are correct? Pause while you choose and press play when you're ready for answers.

Okay, let's take a look.

Firstly, this angle must be a right-angle, angle ABO, because it's a tangent meeting a radius and they always meet at a right-angle.

Therefore, statements A, D, E, and F are all true.

Okay, it's over to you now for Task B! This task contains three questions and here is question one.

Pause while you do this and press play when you're ready for more questions.

And here are questions two and three.

Pause while you do these and press play when you're ready for answers.

Okay, let's go through some answers.

For question one, you had some missing angles to find.

Angle AOB is 55 degrees and here's your justification for why.

Angle CDE is equal to 53 degrees and here's your justification this time.

Angle GHF is equal to 68 degrees and here's your justification.

Angle IFG is equal to 68 degrees as well and here's your justification for that.

Then question two.

You had to prove that the tangents at either end of the diameter are parallel to each other.

Well, you do this by acknowledging that the diameter AB can be broken into two radii, radius OA and radius OB.

Each of those radii are perpendicular to the tangent it meets.

So angle OAC and angle OBD are both 90 degrees because the tangent at any point on a circle is perpendicular to the radius at that point.

And we can see that those two angles are co-interior between a pair of lines.

And when co-interior angles between a pair of lines sum to 180 degrees, like they do here, it means those lines must be parallel.

Therefore, those two tangents are parallel.

Then in question three, you were given that the area of the circle is 144 pi square units, and you had to find the area of the triangle ODE and give your answer in exact form.

This is quite a problem.

The best way to solve this is by working out as much information as you can.

Work out as many lengths as you can in particular.

So if the area of the circle is 144 pi square units, then the radius of the circle must be 12 units because 12 squared times pi is 144 pi.

We also know that we have a tangent that meets a radius at a point on the circle and the angle between those must be 90 degrees.

So angle OBA is 90 degrees.

Also, angle OFD is 90 degrees as well.

And the reason for that is that CA and ED are parallel to each other.

Therefore, corresponding angles on a pair of parallel lines are equal.

We could also look at the length of OA, which is made up of OD and DA.

OD is a radius of the circle, so that's 12 units.

DA is given to us as six units.

So length of OA must be 18 units.

And then the length of OD and OB are equal.

They're both 12 units.

And then we can work out the length of OF by using similar triangles.

The scale factor from the large triangle OBA to the smaller triangle OFD is 12 eighteenths.

So we do 12 eighteenths multiplied by 12 to get eight units.

And then since the triangle OFD is a right-angled triangle, we can use Pythagoras' theorem to work out the missing length FD.

If you do that, you get four root five.

If FD is four root five, then ED must be eight root five.

That's because OB bisects the chord ED.

And then once you've got that, you've got all the information you need to find the area of the triangle ODE, using the formula half times base times perpendicular height.

And if you do that, you get 32 root five square units.

If you got that, a big well done.

Fantastic work today! Now let's summarise what we've learned.

A tangent to a circle is a line, a ray, or a line segment that touched the circle exactly once on its circumference.

We can calculate angles of polygons, where some of the sides of the polygon are tangents to the circle.

And we can consider chords of the circle which are parallel to this tangent.

The circle theorem is, the tangent at any point on a circle is parallel to the radius at that point.

That's the circle theorem we've been focusing on in today's lesson.

And we can demonstrate this theorem by considering a special case of the theorem that the perpendicular from the centre of a circle to a chord bisects the chord.

Great job today.

Have a great day.