Lesson video

Hello, my name is Dr.

Rowlandson and we have a great lesson in store for today.

So let's get started.

Welcome to today's lesson from the unit of circle theorems. This lesson is called: the tangents from an external point are equal in length.

And that's a circle theorem, which by the end of today's lesson, we'll 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 need to remind yourself what any of these words mean, and then press play when you're ready to continue.

The lesson is broken into two learning cycles and I'm going to start with investigating angles and tangents.

Let's begin by reminding ourselves some key properties about tangents, and in particular, the circle theorem that states the tangent at any point on the circle is perpendicular to the radius at that point, and that's what we can see in this diagram here.

We have a tangent meeting a radius, and the angle between them is 90 degrees.

Now this is true for every tangent and radius that intersect at the circumference of the circle, and we can see that with the other examples now on the screen.

Let's check what we've learned with that.

Here we've got a diagram containing tangents and some other line segments.

Could you please work out the size of angles DCE, DEC, and OBA? 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 DCE is 90 degrees.

That's because the angle between a tangent and a radius at a point on a circumference is always 90 degrees.

And then, the other two angles could be worked out by using the fact that angles and triangles sum to 180 degrees.

Angle DEC is 28 degrees.

Angle OBA is 37 degrees.

So here we have a diagram with a circle, its radius, and a tangent which meets at 90 degrees.

It is possible for two tangents to intersect at a point outside the circle.

For example, if we have another tangent here, yes, this doesn't quite intersect yet, but if we extend it, we can see that those two tangents now intersect at a point outside the circle.

If the two tangents intersect, a quadrilateral is created that has two interior right angles.

In this case the quadrilateral is OABC.

We can see it has two exterior right angles, but also it has two interior right angles as well.

Let's check what we've learned with that.

Here we've got a circle with two radii and two tangents marked.

Could you please work out the size of angles OCB and ACB? Pause while you do it and press play for answers.

Let's take a look at some answers.

Angle OCB is 90 degrees.

That's because a tangent meets radius at a point on a circle at 90 degrees.

That happens twice within this diagram.

Therefore, angle ABC will be 50 degrees because angles inside a quadrilateral sum to 360 degrees.

This time we have two diagrams, each of a circle containing points A and C in the same positions.

Point O is the centre of the circle.

Alex says, "Angle ABC is the same as angle ADC because they are both angles subtended by the same chord AC." Now this is incorrect.

Explain why this statement is incorrect.

Pause the video while you write some stuff down and press play when you're ready for answers.

Well, one way we could explain why these two angles are not equal to each other is by working out what they are.

So let's do that.

With the diagram on the left, AB and BC are both tangents.

And the tangent at any point on a circle is perpendicular to the radius at that point.

So we have right angles.

In particular, angle OAB and OCB are both 90 degrees.

And now we have that, we can use the fact that angles in a quadrilateral sum to 360 degrees to work out that angle ABC is 35 degrees.

So what about the diagram on the right? Well, AD and DC are not tangents.

Therefore, OA and AD cannot be perpendicular to each other, so we need a different approach there.

So reflex angle AOC is equal to 215 degrees and the angle at the centre of the circle is twice the angle at any point of circumference.

Therefore, the angle ADC would be equal to 107.

5 degrees, not 35 degrees.

For any point in any location outside the circle, it is possible to find exactly two tangents to the circle that pass through that point.

For example, here we have a circle and there's a point outside the circle marked N.

Where are the two tangents to this circle that pass through point N? I'm going to show in a moment, but perhaps pause the video and try and visualise where those tangents would go.

And then, press play when you're ready to see.

Let's take a look.

One tangent would go here and the other would go here, and each of these tangents are perpendicular to radius at that point.

So let's check what we've learned with that.

Here we've got a circle drawn on a square grid and one of the two possible tangents that passes through point T is drawn.

Could you please find the length of AT? Pause the video while you do that and press play for an answer.

The answer is 15 units.

You can get that by counting the squares.

Here is the other possible tangent that can go through T.

This time, could you use Pythagoras' theorem to find the length of BT? Pause the video while you do that and press play when you're ready for answers.

The answer is also 15 units.

You can get that by forming a right-angle triangle.

You'd have a height of 9 units, a base of 12 units, and 9 squared plus 12 squared is h squared.

And then, you can simplify that to get 15 units.

Okay, it's up to you now for task A.

This task contains four questions and here are questions 1 and 2.

Pause the video while you work through these and press play when you're ready for more questions.

And here is question 3, pause the video while you do this and press play when you're ready for question 4.

And here is question 4, pause the video while you work through this and press play when you're ready for answers.

Okay, let's take a look at some answers.

For question 1, the value of x is 107.

The angle which is labelled LON, which is x degrees, is 107 degrees.

The way you get to that is by first showing that you have two right angles in that quadrilateral because a radius meets a tangent at a right angle.

And then, you can use the fact that the interior angles of a quadrilateral sum to 360 degrees to get 107 degrees.

Question 2, you're given that angle COD is 66 degrees and need to use that fact to eventually calculate the size of angle CAB.

Write an your explanation along the way.

Well, if you start with COD being 66 degrees, COB will be 114 degrees because they form a straight line segment and the angles would sum to 180 degrees.

And then once you have that, you can then look at elsewhere in the diagram.

You have two cases where radius meets a tangent, and therefore you've got two more angles which are both 90 degrees.

That's angle ACO and angle ABO, and you have a quadrilateral whose angles will sum to 360 degrees.

So angle CAB would be 360 degrees, subtract 90 degrees, subtract 90 degrees, subtract 140 degrees, and that gives 66 degrees.

Then on question 3, you're given a circle and a tangent and the tangent goes through point P.

And you had to use your ruler to draw a second tangent through P and you can see that on the screen now.

And then, you have to use a pair compasses to set it to the length of PT and draw a circle through the centre at P.

Now what you can see on the screen here is an arc of that circle.

You may have drawn the full circle.

For part c, you have to write down any observations you made about the circle you constructed.

where the circle or arc passes through the two points where the two tangents intersect the circle.

And then, you have to use a ruler to measure the length from P to the point at the tangent intersects the circle each side, and both tangents should measure to the same length regardless of how large or small it was on your working diagram.

Then on question 4, you had to draw on two tangents through point P.

You can see them on the screen here.

And then, you had to use Pythagoras' theorem to calculate the length of AP and BP.

If you did that, you should get 13 units for each.

And then, you have to find the perimeter of the quadrilateral PAOB.

Now, the radius is 13 units.

That's not easy to see when you go straight from O to B or O to A.

But by looking vertically or horizontally, you can see that the radius is 13 units and that'll be the same for OB and OA.

Once you use that and you can use the answers you found in part b, you can get the perimeter as 52 units.

You're doing great so far.

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

Here we have a circle of a point outside the circle marked P, and here we have Alex who thinks he's spotted something.

He says, "If you draw one tangent from a point outside the circle and measure distance PT1," and by PT1, he means he's marked the point at the circle T1, presumably he's gonna have T2 somewhere else on there.

So the length PT1 is the length from P to the point T1.

So he says, "If you draw one tangent from a point outside the circle and measure distance PT1, and then draw the other tangent from the same point and measure the distance PT2, then the distance PT1 will always be equal to the distance of PT2." Ah, he says, "I can check this using a pair of compasses and drawing an arc with the same distance.

The arc will always pass through PT1 and PT2." So this brings us to our circle theorem of today's lesson and that is the tangents from any external points are always equal in length and that's what we can see on this diagram here.

The distance from P to each of those two points where the tangents meet the circle, T1 and T2, those distances are always equal.

So let's check what we've learned.

Here we've got a diagram of a circle, two radii, and two tangents that meet at a point outside the circle.

What is the length of the line segment PB? Pause the video while you write it down and press play for an answer.

The answer is 105 centimetres and that's because the distance from P to A will be the same as the distance from P to B.

Here we have another diagram and you've got a set of statements.

Which of these statements are always correct for the two tangents that intersect at point P? Pause the video while you choose and press play for answers.

The statements that are correct are b, d, and e.

PA is equal to PB because the tangents from an external point are always equal in length.

AO is equal to OB because they are both radii.

And angle PAO is equal to angle PBO because the radius and tangents are perpendicular at a point on a circle.

Let's check with one more question.

PA and PB are tangents to the circle and the following statements are correct about this diagram.

PA is equal to PB because they are both tangents from an external point.

AO is equal to OB because they are both radii.

And angle PAO is equal to PBO because they're both 90 degrees.

Which types of quadrilateral could AOBP be? Pause the video while you think about this and write down some answers and press play when you're ready to see some answers.

The quadrilateral could be a kite because a kite has at least one pair of equal angles and you can see that PAO is equal to PBO.

The quadrilateral could also be a special case of a kite such as a square and that would be if AO was equal to P in length and that would mean that the angle AOB would also be 90 degrees as well.

If you want to explore this further, there is a link to a GeoGebra file on the bottom of this slide.

So the circle theory that we've been focusing on in today's lesson is that the tangents from an external point are equal in length.

Alex says, "Is there a way of explaining why the two tangents are always equal in length?" Let's look at that together.

Here we've got a diagram with two tangents that meet at point P.

We can show that the length of PA is equal to length of PB by identifying two congruent triangles via the RHS rule for congruence.

That's Right-angle, Hypotenuse, and Side.

If we can do that, we can show that PA is equal to PB.

We can do it by first drawing on these two radii here.

And then, we could split this quadrilateral into two triangles by drawing the line segment from P to O.

Now we have two triangles.

The aim is to show that triangles PAO and PBO are congruent by the RHS rule.

Well, angle PAO is equal to angle PBO when they are both 90 degrees and that's because the tangent at any point in a circle is perpendicular to the radius at that point.

So we've got the right angle.

We could also say that the length of AO and the length of BO are equal because all radii in a circle are equal in length.

So we've got one the sides.

What about the hypotenuse? Well, both triangles share the same hypotenuse that's PO.

That means we have shown that the triangles PAO and ABO both have a right angle.

They both have hypotenuse of equal length and they both have a second side of equal length.

Therefore, by the Right-angle Hypotenuse Side rule, both triangles are congruent.

And if both triangles are congruent, then the third side will also be of equal length to each other as well.

Therefore, we know that PA is equal to PB.

The length of each tangent from the same point P is equal to each other.

Let's check what we've learned.

Here we've got a diagram where PA and PB are tangents to the circle.

You're told that OP is 20 centimetres.

Could you please calculate the length of AP? And leave your answer as a surd.

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

We can do this by using Pythagoras' theorem and that's because we have a tangent that meets a radius and they are perpendicular to each other.

So we have a right-angle triangle OPA.

Therefore, we can do 7 squared plus the length of a AP squared equals 20 squared.

Rearrange it and simplify and we get 3 root 39.

So with that in mind, could you calculate the length of BP? Pause the video while you do it and press play for an answer.

Well, this one's a bit easier.

It is also 3 root 39 and that's because the length of the tangents from external point are equal.

We can use our new circle theorem now to solve problems. Here we have a diagram where we have AO and BO, which are both the radii of the circle, and are both 10 centimetres.

QA and QB are both tangents.

And we're told that the angle ACB, that's the angle at a circumference at towards the top of the diagram that is 60 degrees.

And our job is to find the area of the kite AOBQ and give our answer to three significant figures.

We're going to have to break this down into lots of small steps.

Perhaps pause the video and think about what any of those steps might be.

Is there any information you could put on this diagram that might help make this a little bit easier or even just take us a little step towards the solution? Pause the video while you think about it and press play when you're ready to continue together.

Let's start by working out some missing information.

We know that angle ACB is 60 degrees, so we can work out another angle.

We could say that angle AOB is 120 degrees because the angle at the centre is twice the angle at any point on a circumference.

So now we have at least an angle inside that kite that might be useful.

You may not necessarily know how to find the area of a kite, but what you do know is how to find the area of a triangle and we could split this kite into two congruent triangles.

The line segment QO bisects the angle AOB.

That's because a kite is symmetrical.

Therefore, the angle AOQ would be 60 degrees.

So if we could work out the area of one of these triangles, we'd only need to double it to find the area of the kite.

Well, these triangles are right-angle triangles and that's because the tangent at any point on a circle is perpendicular to a radius at that point.

The fact that these triangles have two perpendicular sides means that if we can find the lengths of those sides, we can use the formula 1/2 times base times height to find the area of each triangle.

So how can we get the lengths of those two perpendicular sides? Well, we were given one of them.

AO was 10 centimetres.

So if AO was 10 centimetres, then the length AQ can be found by using trigonometry.

We have a right-angle triangle.

We have one length and an angle.

That is great for using trigonometry.

So AQ would be equal to 10 multiplied by 10 of 60 degrees and that would give 17.

32050 and more decimals.

Now if you want to, you could write that number down and use it or it might be more accurate to use 10 tan 60 in any future calculations instead.

So we continue using all of the possible decimals.

So to find the area of the triangle AOQ, we can do 1/2 times 10 tan 60 times by 10, which is the length of the radius AO, and that'll give 86.

6025 and so on.

And then, we could multiply it by 2 because triangle BOQ is congruent to AOQ, which means the quadrilateral is twice the area of the triangle and the area of the quadrilateral would be 173.

2050.

So let's check what we've learned.

Here you've got a kite BOAQ.

You're told that the angle BQA is 44 degrees and the length the radius is 9 centimetres.

Could you please calculate the area of the kite BOAQ? And give your answer to three significant figures.

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

Well, if you first split this kite into two congruent right-angle triangles, you'll see that angle BQO is 22 degrees.

And then, you could do BQ multiply by tan 22 to get 9, and rearrange it to get the length of BQ as being this: 9 over tan 22.

And then, you could use that with the formula 1/2 times base times height to get the air of the triangle as 1/2 times 9 times 9 over tan 22.

And that will give you 100.

24 or more decimals.

So the area of the kite will be double that which give you 200 square units to three significant figures.

Okay, it's up to you now for task B.

This task contains four questions and here is question 1, pause the video while you do it and press play for question 2.

And here is question 2, pause the video while you do this and press play when you're ready for question 3.

And here is question 3, pause the video while you do this and press play for question 4.

And here is question 4, pause the video while you do this and press play for some answers.

Okay, let's take a look at some answers.

Here are the answers to question 1, pause the video while you check this against your own and press play for more answers.

And here is the answer to question 2a, pause the video while check this against your own and press play when you're ready for more answers.

Here is the answer to question 2b.

Pause while you check this against your own and press play for more answers.

Here's the answer to question 3a, pause while you check and press play to continue.

And here's the answer to question 3b, pause while you check and press play to continue.

Here are the answers to questions 4a and 4b, pause while you check these and press play to continue.

And here's the answer to 4c, pause while you check this and press play for today's summary.

Fantastic work today.

Now let's summarise what we've learned.

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

We've also learned that exactly two tangents pass through any point outside of a circle, which also means that any two non-parallel tangents will always intersect at a point outside of the circle.

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

We've seen examples of that with triangles and quadrilaterals.

Two tangents from an external point are always equal in length.

And we can start thinking about kites in that situation because a kite is formed from two radii and two tangents that meet at a point.

Well done today, have a great day.