Lesson video

Hello, Mr. Robson here.

Welcome to maths.

We're finding the equation of a radius of a circle today.

Well, that sounds really cool.

Let's find out a bit more about it.

Our learning outcome is that we'll be able to find the equation of a radius of a circle.

What's keywords in today's lesson? One of which is radius.

The radius is any line segment that joins the centre of a circle to its edge.

Another key word is gradient.

The gradient is a measure of how steep a line is.

It is calculated by finding the rate of change in the y-direction with respect to the positive x-direction.

We've got three parts to our learning today.

We're going to begin with a bit of exploration.

We're gonna explore the centre of a circle.

We'll use graphing technology to explore the centre of a circle, so go to Desmos.

com and click Graphing Calculator.

Your screen should look just like mine once you've done that.

Pause and make sure it does.

Once you're in Desmos, what we're gonna do is graph some equations.

I'd like you to graph these four equations and leave them on the same axes.

You want to scroll and zoom so that your axes look about the same as mine.

Pause and graph these four equations now.

Welcome back, hopefully, you've got these four circles.

We've got concentric circles with radii of one, two, three, and four respectively.

This is because the general form of a circle is x squared plus y squared equals r squared with r, of course, standing for radius.

Quick check, you've got this.

I'd like you to graph a circle with radius eight centred at the origin.

What equation do you think you're going to type in to do that? Pause, see if you can figure it out.

Welcome back, hopefully, you got this.

That's the graph of x squared plus y squared equals eight squared.

Remember that general form, x squared plus y squared equals r squared.

If you want the radius of eight, you have to type eight squared on the right-hand side of that equation.

So will all graphs of a circle be centred at the origin? Well, all the ones we've seen so far are.

Let's explore this.

What I'd like you to do is type in this equation, (x minus a) squared plus y squared equals two squared.

Once you type that in, Desmos will invite you to add a slider.

Click on add slider.

Once you've done that, your screen should look like this.

Then I'd like you to hit that play button and observe what changes.

Pause, make sure your screen looks like mine and hit that play button now.

Welcome back, I wonder what you observed.

Hopefully, you noticed our circle maintained the same radius, but its location changed.

The center's location was somewhere on the x axis.

You saw that circle travel left and right along the x axis, didn't you? To explore those centres more closely, we can change the interval and the steps on our slider.

What do I mean by that? If you click on one of the numbers at either end of your slider, this will come up.

From here, we're gonna change the interval.

I'd like you to change the interval so it reads -5 is less than or equal to a is less than or equal to +5.

And I'd also like you to enter a step of one.

Your screen should look like that.

Now, I'd like you to hit play and observe what changes here.

Pause and do that now.

Welcome back, I hope it was easier to see what was changing when we had the slider set like that.

You would've noticed that when the value of a was equal to one, the centre of the circle was at (1,0).

And when a equaled two, the centre of the circle was at (2,0).

So quick check, did you correctly observe where was the centre when a equaled three, when a equals four, when a equals five? I'd like you to pause and fill in those three blank coordinate pairs.

Welcome back, hopefully, you said when a equals three, the centre is at (3,0).

When a equals four, the centre is at (4,0).

And you know what's coming next, don't you? When a equals five, the centre is at (5,0), well done.

So how about when a was negative? Well, I observed that when a was equal to -1, the centre of the circle was at (-1,0).

Let's see if you make the right observations about when a equals -2, -3, -4, -5.

I'd like you to pause and fill in those blank coordinate pairs.

Welcome back, hopefully, you had (-2,0), (-3,0), (-4,0), (-5,0).

What we're going to do now is put all those observations together and see if we can make a generalisation.

On the left-hand side there, you see when a equals one, the centre of the circle is at (1,0).

When a equals two, the centre is at (2,0).

All are positive a integer values.

And on the right-hand side, on negative integer values, when a equaled -1, the centre was at (-1,0).

When a equals -2, the centre was at (-2,0) and so on.

From all this information that we've got, can we now complete the generalisation at the bottom of the screen? The circle (x minus a) squared plus y squared equals r squared has a centre at? Well done, a zero.

By changing the value of a, we were able to change the x-coordinates of the centre of our circle.

This generalisation is important.

You want to write it down.

Pause and do that now.

Quick check, you can apply that generalisation now.

I'd like you to fill in these blanks.

The equation x minus 24 squared plus y squared equals 36 is a centre with circle of what coordinate and a radius of what value? For the second sentence, I'm giving you the coordinate of the centre and the value of the radius and I'd like you to put it back into an equation.

Pause, have a think about this.

See if you can fill in those blanks.

Welcome back, hopefully, you said for the equation x minus 24 squared plus y squared equals 36, that circle has a centre at (24,0).

And if r squared is equal to 36, then r must be six.

For the second sentence, a circle with centre (-30,0) must be x plus 30 squared plus y squared with a radius of 10.

The right-hand side will be 10 squared, that's a 100.

You might have written that second equation like so, but that would be a way less efficient way of writing it.

So will all graphs of a circle have a centre on the x axis? Let's explore that.

What I'd like you to do is type in the equation, x squared plus y minus b squared equals five squared.

Desmos will again invite you to add a slider this time to change the value of b.

Once you've added your slider, could you set the interval and the step to match mine? Once you've done that, press play.

Let's see what changes this time.

Pause and do this now.

Welcome back, hopefully, you noticed that our circle maintained the same radius, but this time, the centre changed where it was located on the y axis.

You saw that circle travel up and down the y axis.

Hopefully, you observed that when b equaled one, the centre of the circle was (0,1).

And when b equaled two, the centre of the circle was (0,2).

Quick check, can you spot it where the centre of the circle was when b equaled three, b equaled four, and b equaled five? I'd like you to pause and fill in those three missing coordinate pairs.

Do this now.

Welcome back, hopefully, you noticed when b equal three, the centre was at (0,3).

When b equaled four, the centre was at (0,4).

When b equals five, the center's at (0,5).

What about when b had a negative integer value? When b equals -1, the centre of the circle was (0,-1).

When b equaled -2, the centre of the circle was (0,-2).

Can you pause now and fill in those three missing coordinate pairs for b equals -3, -4, and -5? Do this now.

Welcome back, hopefully, you said when b equals -3, the centre was at (0,-3).

When b equals -4, the center's at (0,-4).

And when b equals -5, the center's at (0,-5).

Again, let's put all those results together and see if we can come up with a generalisation.

Look at the bottom of the screen.

The circle x squared plus y minus b squared equals r squared as a centre at? Well done, (0,b).

Again, this generalisation is really important.

You'll want to pause and write that down.

Let's see if we can apply that generalisation now.

I'd like you to fill in the blanks of these two sentences.

For the equation x squared plus y minus 12 squared equals nine, that'll be a circle with centre, what coordinate pair, and radius, what length? And then secondly, a circle with centre (0,-7) and radius eight will have the equation of what? Pause, fill in those blanks now.

Welcome back, hopefully, you said for x squared plus y minus 12 squared, we'll have a centre of (0,12).

And if r squared equals nine, then r itself must be three.

Secondly, our circle with centre (0,-7) must be x squared plus y plus seven squared.

And if we've got a radius of eight, r squared will be 64 on the right-hand side.

You might have written this as x squared plus y minus -7 squared equals eight squared, but it's way less efficient way to write it.

Let's see if we can combine all our learnings so far.

What can you deduce about a circle with this equation? x minus a squared plus y minus b squared equals r squared.

How did the value of a affect the location of the centre of our circle? How did the value of b affect the location of the centre of our circle? Well, we found that when a was positive, our centre moved in a positive x direction.

We found that when b was positive, our centre moved in a positive y direction.

So where do you think the centre of this circle will be? I'd like you to pause and have a think about this.

And don't worry, I'll be back in a few moments with the answer.

Welcome back.

Hopefully, you said the centre of our circle will be x equals +4 and y equals +7.

That would give us the coordinate pair (4,7).

The centre of this circle will be at (4,7).

It's really important when we make a conjecture like this that we use tech to check.

Let's type it into Desmos and see if that's right.

Wonderful, there it is, circle with a centre at (4,7).

Equations in the form x minus a squared plus y minus b squared equals r squared have a centre at (a,b).

This generalisation is really important.

You'll absolutely want to pause and write that down.

Quick check, you've got that.

Using that generalisation, can you match these equations to the centre of their circle? There's five equations on the left-hand side.

Five coordinate pairs on the right-hand side.

Which centre belongs to which circle? Pause, match them up now.

Welcome back, you would've noticed that all the radii are the same in those equations so it's all about the left-hand side.

(x minus nine) squared, (y minus two) squared, the centre of that circle will be at (9,2).

(x minus two) squared, (y plus nine) squared, centre of that circle will be at (2,-9).

(x plus three) squared, (y minus six) squared, centre of that one will be at (-3,6).

So we must be left with (x plus two) squared plus (y plus nine) squared being at centre (-2,-9).

And finally, a centre of (-6,3) for that bottom equation.

Andeep is considering the graph of this circle, (x minus five) squared plus (y plus five) squared equals one.

Andeep says, the centre of this circle will be at (-5,5).

Well, that makes sense, doesn't it? I see a -5.

I see a +5.

So the centre will be at (-5,5), wouldn't it? Andeep has made a common error that we frequently see when examining the centre of circles.

I'd like you to write a sentence to explain to him why he's wrong.

Pause and do this now.

Welcome back, you might have written equations in the form (x minus a) squared plus (y minus b) squared equals r squared have a centre at (a,b).

Try to compare the equation to this form.

If you write the generalisation next to our original equation, you can very easily see that our a value must be five, our b value must be -5, so the centre must be at (+5,-5), not the way round that Andeep had it.

Pay particular attention to this slide because this is a really common error.

Practise time now.

For question one, I'd like to complete these statements.

There's the equation of a circle.

For part A, I'd like to know the coordinate pair of its centre and the length of its radius.

Something very similar for part B as a missing coordinate pair and a missing radius.

For part C, you have the coordinate pair of the centre.

You have the length of the radius.

I'd like to know the equation.

Pause and do these three now.

For question two, here's two circles.

I'd like to know their equations.

Pause and write those two equations now.

For question three, I like this one.

I'd like you to recreate this image using graphing technology.

Tricky problem, but a fun one.

I'm sure you'll agree.

Pause and give it a go now.

Welcome back, feedback time.

For question one, we were completing these statements.

For part A, we should have had a centre at (3,5) and a radius of four.

For part B, we should have had a centre at (9,-13) and a radius of nine.

For part C, we had the centre.

We had the radius.

We're putting it into an equation.

Your equation should have read (x plus one) squared plus (y plus seven) squared equals a 144.

For question two, I asked you for the equations of these two circles.

If we want to do this, we wanna identify the centre and the length of the radius.

From there, we can put it into our generalisation.

That must be the circle (x plus five) squared plus (y minus four) squared equals four squared, that's 16.

For the second circle, centre at (4,-2) and a radius of five should give you that equation.

(x minus four) squared plus (y plus two) squared equals 25.

Question three, I hope you thoroughly enjoyed doing this one.

You needed those six equations to recreate this image.

You might wanna pause and just check out your equations match mine.

Onto the second part of our learning for today, we're gonna look at the gradient of the radius.

If we know the centre of the circle and the coordinates of a point on the edge, we can calculate the gradient of the radius that connects them.

For example, we know the centre of this circle is (6,5).

We have a point on the edge (9,9).

There's a few ways we could do this.

We could draw a right triangle showing the change in y and the change in the positive x-direction and we can declare that the gradient is four over three.

You'll often see gradient written as m in mathematics.

So we'd say m equals four over three for this radius.

We could have used the formula for gradient.

Gradient is the second y-coordinate minus the first y-coordinate over the second x-coordinate minus the first x-coordinate.

Plugging in, in this example would've seen us do nine minus five for the y-coordinates and the enumerator position and nine minus six for the x-coordinates in the (indistinct) position, giving us four over three.

Using the formula can often be the quickest method when we've not got a visual representation.

If we had this example, the equation of a circle is (x minus eight) squared plus (y plus two) squared equals 50 and a point on the circumference is (13,-7).

Find the gradient of the radius.

Well, we know from the equation of that circle that the centre is (8,-2).

Then just plug into the formula for gradient.

We'll get -5 over five which gives us a gradient of -1.

With a problem like this, a sketch or graphing technology allows us to check the validity of our answer.

When we type the equation of that circle into Desmos, we do indeed see the centre at (8,-2).

We can see that point on the circumference at (13,-7).

The gradient of the radius connecting those two is indeed -1.

Quick check, you've got this.

The equation of a circle is (x plus seven) squared plus (y minus one) squared equals a 100 and a point in the circumference is (-15,-5).

I'd like you to find the gradient of the radius.

Once you've done that, I'd like you to use tech to check your answer.

Pause, give this a go now.

Welcome back, hopefully, the first thing you did was identify the centre of that circle (-7,1), and then plugging into our formula for gradient.

We would've found gradient of six over eight which, of course, we cancelled down to three over four.

And when we type the equation of the circle and that point (-15,-5) into Desmos, we find that tech validates our answer, a gradients of three over four for that radius.

We won't always be given the circle in the form (x minus a) squared plus (y minus b) squared equals r squared.

This example looks a little trickier.

The equation of a circle is x squared plus y squared equals six x minus two y plus a 159 and a point on the circumference is (-9,4).

Find the gradient of the radius.

It's unusual this form, but we can rearrange it that it's in a form we're familiar with.

If we put all the variable terms on the left-hand side, what we can then do is turn x squared minus six x into x minus three squared.

But crucially, you'll notice because (x minus three) squared is x squared minus six x plus nine.

By rewriting it like this, we've added nine to both sides, hence 159 plus nine.

We've got 168 on the right-hand side now.

Let's do something similar for y squared plus 2y.

We'd rewrite that as y plus one squared, but of course, we've added one to both sides so we added one to the right-hand side as well.

Now that we've got it in this form, we know that the centre of this circle is at (3,-1).

If we know the center's at (3,-1) and we know there's a point in the circumference (-9,4), we can just plug that into the formula for gradient and we will find a gradient of -5 over 12.

Quick check, you've got that.

The equation of a circle is x squared plus y squared equals 10y minus 18x minus 88 and a point on the circumference is (-6,8).

I'd like to find the gradient of the radius.

Pause, give this a go now.

Welcome back, hopefully, you started rearranging to get to the form that we want to work in.

Putting all the variable terms on the left-hand side.

And when we put x squared plus 18x into the brackets x plus nine squared, we've actually added 81 to both sides.

When we do something similar with the y squared and y terms, we've added 25 to both sides.

But crucially, in that form, we can see the centre of the circle (-9,5).

Plugging our two coordinate pairs into the formula for gradient, we get a gradient of three over three IE one.

When we tap the equation at circle into Desmos, we do indeed find that we're correct.

It's always useful to use tech to check.

Practise time now.

For question one, this circle has a centre (-5,-3).

For part A, I'd like to find the gradient of the radius which connects that centre to the point (-10,-15).

And for part B, the radius connecting the centre to the point (7,-8) on its circumference.

Pause and work these two out now.

Question two, the equation of a circle is (x minus 13) squared plus (y plus 31) squared equals 625.

I'd like you to find the gradient of the radius which connects the centre to the point (6,-55) that lies on its circumference.

Question three is a tricky one, but we'll give it a go.

The equation of a circle is x squared plus y squared equals 24y minus 20x plus 45.

That's gonna require some rearrangement.

I'd like to find the gradient, the radius which connects the centre of the circle to the point (5,4) that lies on the circumference.

Pause, give these two a go now.

Welcome back, feedback time.

Let's see how we got on.

Question one, part A, we should have found the gradient of that radius to be 12 over five.

For part B, we should have found the gradient of that radius to be -5 over 12.

For question two, we should have identified that the centre of the circle is at (13,-31).

From there, we can plug our two known coordinate pairs into our formula for gradient.

Find the gradient to be 24 over seven.

With question three, we would've started by rearranging the equation.

Rearranging it like so gets it into a format which we want to work with.

Once we're in that form, we know the centre is at (-10,12).

We can now plug our two coordinate pairs into a formula for gradient and we find the gradient is -8 over 15.

Onto the third and final part of our learning now.

It's quite a short on this.

We've already done all the hard work.

This final little piece of the jigsaw is the equation of the radius.

If we know the equation of the circle and the coordinates of a point in the circumference, we can find the equation of that radius.

The equation of a circle is (x minus seven) squared plus (y minus 10) squared equals a 169.

A point on the circumference is (19,15).

Find the equation of the radius.

Well, the equation of the radius is gonna be a linear graph of the form y equals mx plus c.

Step one, find the gradient.

We know how to do that.

We know the centre.

We've got a point in the circumference.

We can plug it into the formula for gradient.

We'll find the gradient is five over 12.

Step two, we wanna find the constant in y equals mx plus c.

So we've got a linear graph of y equals five over 12x plus c.

We know the centre of this circle is (7,10), so we can plug in x equals seven, y equals 10 into that equation.

We find the value of c to be 85 over 12.

Therefore, the equation of the radius must be y equals five over 12x plus 85 over 12.

It's always useful in math to check our work.

In this case, we can use a second coordinate to check that that is indeed the equation of our line.

If we substitute in, x equals 19, y equals 15, they do indeed satisfy the equation so we know we're right.

We can also use tech to check.

You type in the equation of that circle, equation of our radius and you see this.

Our radius is a segment of that linear graph.

Quick check, you can repeat that scale.

The equation of a circle is x minus 50 squared plus y minus 75 squared equals 1,681.

A point on the circumference is (10,84).

I'd like you to find the equation of the radius.

Pause and do this now.

Welcome back, hopefully, you spotted the centre of that circle is at (50,75).

Once we know the centre of the circle, we've got the point in the circumference, find the gradients.

Our gradient would be -9 over 40.

Let's find the constant in y equal mx plus c.

Substituting in the known coordinate pair (10,84), we'll find c to be 345 over four.

Therefore, the equation of our radius is y equals -9 over 40x plus 345 over four.

Always useful to use a second coordinate to check our line.

When we do that, we find the equation is satisfied.

We must be right.

Additionally sensible to use tech to check.

When you type all this in, you find that our radius is indeed a segment of that linear graph, well done.

Practise time now.

Question one, the equation of a circle is (x minus five) squared plus y squared equals 50.

A point on the circumference is (0,5).

I'd like you to find the equation of the radius.

Crucially, I'd like you to check your answer so you know you're correct.

Question two, the equation of a circle is (x plus 20) squared plus (y minus 12) squared equals 25.

A point on the circumference is (-16,15).

I'd like you to find the equation of the radius.

And once again, check your answer.

Pause, do these two now.

Welcome back, let's see how we go on.

Hopefully, you identified the centre of this circle is (5,0).

A gradient will be -1 on this occasion.

The constant c was equal to five.

Therefore, the equation of the radius y equals five minus x.

Check one is to substitute in the second end coordinate and we know we're right.

Check two, use tech to check.

For question two, the centre of the circle was at (-20,12).

The gradient, three over four.

Constant c is 27.

Therefore, the equation of the radius is y equals 3 over 4x plus 27.

We can check by substituting in the second coordinate and we know we're right.

You can also use tech to check.

Sadly, we're at the end of the lesson now, but we've learned from the equation of a circle, we can identify the centre of the circle.

And when given both a point on the circumference and the centre of that circle, we can find the gradient and equation for that radius.

Hope you've enjoyed today's lesson as much as I have and I look forward to seeing you again soon for more maths.

Goodbye for now.