Lesson video

Hello, Mr. Robson here.

Welcome to maths, super to see you again.

Today, we're learning about the equation of a circle and that is a pretty awesome bit of maths, so let's get learning.

A learning outcome is that we'll be able to generate coordinate pairs using the equation of a circle in the form x squared plus y squared equals r squared, and then draw the graph.

Keyword that you'll hear a lot today is radius.

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

Two parts to our learning.

We're gonna begin by looking at the equation of a circle.

This diagram shows a quarter of a circle on a coordinate grid.

The centre of the circle is the origin.

If we pick a point on the circumference of the circle, do we have enough information to calculate the radius? What do you think? Well done.

The answer's no, not yet.

What information would we need? Good idea.

A coordinate pair would do it.

If we knew that pair of coordinates, we'd absolutely be able to calculate the radius.

How? Excellent idea.

If we knew a coordinate pair, we could use Pythagoras theorem to calculate the radius.

In this example, we take our x coordinate, square it, our y coordinate, square it, and that would give us a radius squared.

We could use Pythagoras theorem to calculate the radius.

Could we do this with any coordinate pair to find that radius, instead of my first coordinate pair, could I use the second coordinate pair? What do you think? Well done.

Of course we can.

The triangle would look different and the equation would look different, but it's just the x coordinate squared and the y coordinate squared giving us a radius squared.

The diagram now shows a circle on a coordinate grid.

The centre of the circle is still the origin.

Now we know we can use Pythagoras theorem to find the radius if we have any coordinate pair in the first quadrant, but could we use a coordinate pair from any quadrant to find the radius or will the negative values in those coordinate pairs affect the calculation? What do you think? Pause.

Have a conversation with a person next to you or a good think to yourself.

See you in a moment to reveal the answer.

Welcome back.

I wonder what you said.

Hopefully, you said we can use Pythagoras theorem on a coordinate pair in any quadrant to find the radius.

These five examples, we could use those five equations and they'll all work.

Why does having a negative x value or a negative y value in our coordinate pair not matter? Well, it's because squaring makes any negative value positive, negative x squared is the same as x squared.

Negative y squared is the same as y squared.

Say this idea is gonna work in any quadrant.

X squared plus y squared equals r squared because this equation links every point on the circumference to the radius.

It is the general form of an equation of a circle with its centre at the origin.

That equation and that sentence feel important.

You might wanna pause and write them down.

Quick check you've got that.

Which of these is the general form for the equation of a circle centred at the origin? Is it A, B, or C? Pause, tell the person next to you or tell me aloud on the screen.

Welcome back.

Well done.

It absolutely is C and it's not option A.

It is Pythagoras theorem which links the coordinate pairs to the length of the radius.

So in that visual there you can see that x squared plus y squared will indeed equal r squared, hence it's C.

Don't accidentally write x plus y equals r.

They need to be squared.

Why was it not B? Well, some people sometimes think that's the general form for the equation of a circle, but the diameter is double the radius.

The diameter is not the radius squared.

So beware of that misconception.

We can use this general form to draw the graph of a circle.

In this example, x squared plus y squared equals 25.

We have an equation, so we could use a table of values, but this method has problems. This equation is quite a nice example 'cause there isn't plenty of integer values in the table though.

Let's have a look at some of those integer values.

X equals three; in that case three squared plus y squared equals 25; 9 plus y squared equals 25.

So y squared must be equal to 16.

So ask why? Why is positive or negative the root of 16.

So y equals positive or negative four.

Be sure to take both the positive and the negative square roots.

When we put that into our table of values, how nice does that look? We've got coordinate three positive or negative four.

So is that a coordinate pair? No, it's two coordinate pairs.

It's three positive four and three negative four.

X equals four is also gonna give us an integer value for y.

Let's substitute it in, four squared plus y squared equals 25.

So 16 plus y squared equals 25, y squared equals nine.

So y equals three, right? Well done.

Y equals the positive or negative root of nine.

That's positive or negative three.

So we now plot the coordinate pair four, three.

No we don't Mr. Robson, we plot two coordinate pairs.

Four positive three, and four negative three.

Well done you.

Can you imagine what's gonna happen when we make x equal to negative three? Well done.

You remembered.

When we square in, negative becomes positive, so negative three square becomes positive nine and this should look very familiar.

The difference now is our coordinate pairs are negative three four and negative three negative four.

So they appear there on the graph.

We can do the same with x equals negative four.

Negative four squared becomes 16 and this is going to look very familiar and we'll plot negative four, positive three and negative four, negative three on our grid.

At this moment, we have eight identical triangles.

Watch this.

All of these triangles have lengths of three, four, and five.

How nice is that? Eight identical 3, 4, 5 triangles.

When x equals one, it's a bit of a problem when it comes to substitute this one in.

Well, not a problem when we come to substitute it, but the outcome is a problem.

One square plus y squared equals 25.

So y squared must be equal to 24 and 24 is not a perfect square.

So y equals positive or negative.

The root of 24, which is positive or negative, 4.

89897, et cetera.

This is very difficult to plot accurately and we mathematicians, we value our accuracy.

That's where the table of values method falls down when we're plotting the graph of a circle.

When plotting the graph of a circle in this form, it's easiest to start with the intercepts.

The intercepts where the circle cuts the x and y axis, i.

e.

, when x equals zero and when y equals zero.

When x equals zero, well that's the easiest substitution yet y square equals 25, y will be the positive or negative root of 25, it's positive or negative five.

When y equals zero, does this look a little familiar? Aha, when y equals zero, x is going to be positive or negative five.

That's those two coordinate pairs there.

From there, we have the radius of the circle.

We know the centre is the origin.

We can draw our circle and that is the graph of x squared plus y squared equals 25.

We didn't need a table of values and we didn't need to seek non-zero integer values to fit the equation.

If we find the intercepts when x equals zero, when y equals zero, we know the radius.

We can then graph the circle.

In this case, x squared plus y squared equals 25.

The circle has a radius of five, but we knew that from the equation before we even grafted the circle.

How did we know it? Well the general four when centred at the origin is x squared plus y squared equals r squared.

Ah, look at those two equations alongside each other.

R squared must be equal to 25, so r must be five.

You are screaming at the screen now why is it not positive or negative five? I did tell you to take the positive and negative roots, when plotting that table of values.

However, in this case we're calculating the radius, the positive value only this time because the radius is a length, therefore it must be greater than zero.

Your turn now.

I'd like you to find the intercepts to draw the graph of this circle.

X squared plus y squared equals 100.

Where are those intercepts? What's y when x equals zero? What's x when y equals zero? Find those coordinate pairs.

You'll be able to graph this circle.

Pause and do that now.

Welcome back.

Let's see how we did.

Hopefully you substituted in x equals zero to find the y intercepts.

When x equals zero, y will be the positive and negative roots of a hundred that's positive, negative 10.

Pop those coordinate pairs on the graph.

How about when y equals zero? No surprises here.

X squared will be a hundred, x will be positive or negative root of a hundred, which is positive negative 10.

Those two one pairs are there.

We know the centres at the origin, we know the radius.

We can draw the circle like so.

The graph has gone through the point 6, 8.

Show that these values fit the equation.

Thus confirming your graph is correct.

Pause and show why that coordinate pair fits that equation now.

Welcome back.

Hopefully, you substituted in x equals six y equals eight and noticed that six squared plus eight squared does indeed equal 100.

The values satisfy the equation so we know that that must be the correct circle.

It's interesting to compare the two circles we've graphed so far.

X squared plus y squared equals 25 and x squared plus y squared equals a hundred.

When we write those equations in this form, we can see we had a radius of five for the smaller circle and a radius of 10 for the larger circle.

When we compare these two coordinate pairs in the circumferences 3, 4 and 6, 8, we see something rather nice.

Look at that triangle and that triangle.

What do you notice about the two of them? Well done.

They're similar triangles.

We've got two similar triangles with a scale factor of two.

That means our circles are also enlargements with scale factor two.

In fact, all circles centred on the origin are enlargements of each other.

Let's consider this equation again and compare it to another x squared plus y squared equals 25.

We can write it in that form.

X squared plus y squared equals five squared and we can write on the radius five on the circle.

What about this circle? X squared plus y squared plus nine equals five squared.

Alex says circles come in the form x squared plus y squared equals r squared.

So the radius of this one must be five.

Can you see what Alex has done wrong? Pause.

Have a think about that now.

Welcome back.

I hope you said that that equation is not in the form x squared plus y squared equals r squared.

So Alex was wrong to just look at the right hand side and say five squared, if that's r squared, the radius is five.

No, we need to rearrange to get it in the correct form.

We need the constant values on the right hand side, so we'll add negative nine to both sides.

The left hand side now reads correctly, x squared plus y squared.

On the right hand side we've got five squared minus nine, which would be 16.

From there, we can call 16 four squared.

Alex has spotted it now; the radius is four.

Well done Alex.

Let's check.

You can do that.

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

Pause and give that a go now.

Welcome back.

Hopefully you identified we had a constant on the left, which we wanted to get rid of.

So add negative 25 to both sides of the equation.

The left hand side is now correct x squared plus y squared.

On the right hand side, we've got 13 squared minus 25, which is 144.

You recognise that, it's perfect squared, it's 12 squared.

Say we've got x squared plus y squared equals 12 squared.

We've got the correct form.

We can declare that the radius is 12.

Well done.

Practise time now.

Question one.

I'd like you to plot the graphs of these circles.

For a, I'd like the graph of x squared plus y squared equals nine.

And for b, I'd like x squared plus y squared equals nine squared.

Pause.

Give this a go now.

Question two, part a.

Alex plots this graph and we can see some of his calculations.

We can see his table of values.

We can see one of his written calculations.

We can see a calculation he's done in this calculator.

We can see what he's plotted on a grid.

What has Alex done wrong? Write a sentence to explain his error to him and correctly complete the graph.

Pause and do this now.

Question two, part B: Lucas is plotting this graph x squared plus y squared equals four.

Lucas says, I recognise this one immediately.

Equation of a circle.

This graph took me 30 seconds to draw.

Select any coordinate pair to show Lucas why he's wrong and write a sentence to explain how he should have drawn the graph.

Pause and do that now.

Question three, find the radius of these circles.

No need to graph them on this occasion.

Just use the equations to find the radius.

Pause and do this now.

Feedback time now, let's see how we did.

Question one, we were plotting the graphs of circles for part A x squared plus y squared equals nine when x equals zero.

Y was positive or negative three.

That's those two coordinate pairs; when y equal zero x was positive or negative three.

That's those two coordinate pairs.

We can join the circle like so.

For part b, x squared plus y squared equals nine squared, when X equals zero, you is positive or negative nine.

When y equal zero, x is positive or negative nine.

We know the radius centre of the origin.

We can draw the circle.

Question two part eight.

I asked you to decide what Alex had done wrong.

Write a sentence to explain his error and correctly complete the graph.

You might have written, Alex has not taken both the negative and positive roots when calculating values for y.

You can see in his calculated display and his written example, he's only taken the positive roots.

He needs the negative ones, two.

I asked you to draw the graph correctly, in which case you could have reflected those coordinate pairs and completed the graph of the circle like so.

Question two part B, I asked you to select any coordinate pair to show Lucas why he is wrong and write a sentence to explain how we should have drawn the graph.

You didn't have to use this particular coordinate pair.

I've chosen zero four 'cause it's one of the intercepts.

Have you chosen any intercept, that would work beautifully.

By choosing that coordinate pair x equals zero y equals four, I can substitute them into the equation and show that zero square plus four squared does not equal four.

If the x and y values and that coordinate pair do not satisfy the equation, then that must not be the correct graph.

Any intercept would quickly show you why Lucas's graph cannot be right.

You might have used four zero, negative four zero, zero negative four.

The intercepts were the best coordinate pairs to use.

You might then have written this feedback to Lucas substituting x equals zero gives y squared equals four, which tells us that zero two and zero negative two are coordinates.

Therefore the radius is two.

That would've enabled Lucas to draw the graph more accurately.

For question three, asks us to find the radius of these circles, x squared plus y squared equals four cubed.

I hope you didn't write the radius as four.

We need these equations in the form x squared plus y squared equals r squared.

So four cubed is 64.

Oh wonderful.

That's a perfect squared.

It's eight squared.

Therefore the radius of that circle is eight.

B was a little trickier.

I need x squared on the left hand side, so I'm gonna add negative two x squared to both sides.

X squared equals 17 minus y squared.

If I add y squared to both sides, oh well that has worked wonderfully; x squared plus y squared equals 17.

The radius must be the square root of 17.

Onto the second part of our learning now.

Points of intersection.

It is possible to solve this pair of simultaneous equations using substitution.

If we call the first equation equation one, the second equation, equation two, we can rearrange equation two to make y the subject.

Why would we want to do that? Because we can then substitute it into equation one.

Instead of x squared plus y squared on the left hand side, we've now got x squared plus bracket x plus seven squared.

X plus seven squared.

We can expand that, when we expand it, it becomes x squared plus 14 x plus 49.

We can neaten all that up and be left with two x squared plus 14 x plus 24 equals zero.

Common factor of two.

I'll divide every term by two thank you.

X squared plus seven x plus 12 equals zero.

That factor rises and because it factor rises, I know I've got x values of negative three and x equals negative four as solutions to those equations.

I need to substitute those x solutions back in to get the respective y values, and we find these two solutions.

X equals negative three, y equals four, x equals negative four, y equals three.

So for that pair of simultaneous equations, we got this pair of solutions.

x equals negative three y equals four x equals negative four y equals three.

We know that this is the equation of a circle and we know this is a linear equation.

When we graph them, have you spotted something? Well done.

The solutions are the coordinate pairs of the intersections of the two lines.

How nice is that? Solve this pair of simultaneous equations x squared plus y squared equals a hundred and x plus two y equals 20.

We could of course use substitution, but we don't have to use substitution.

We can draw the graphs and find the intersections.

For the graph of x squared plus y squared equals a hundred, well, when x equals zero, y equals positive or negative 10.

That would give us that circle.

For the linear equation, x plus two y equals 20, when x equals zero, y equals 10.

When y equals zero, x equals 20, that's those two coordinate pairs.

It's a linear equation.

Joining with a straight line.

Oh look, we've got some solutions, some intersections.

The solutions of the coordinate pairs, the intersections of the two lines 0, 10.

So we have a solution at x equals zero, y equals 10.

Our next intersection is at 8, 6.

We've got another solution.

x equals eight, y equals six.

That's two intersections.

Therefore two solutions.

We won't always get integer just solutions in this example.

We can still use the intersections to find the solutions, but they'll only be estimates.

In this case, I can spot that intersection there.

An estimated to be at negative six, negative 1.

3, so I've got an estimated solution.

X equals negative 0.

6 y equals negative 1.

3.

My second intersection is approximately at x equals 1.

4 y equals negative 0.

3.

Quick check.

You can do that.

I'd like you to use the graph to estimate the solutions to this pair of simultaneous equations.

Pause and do that now.

Welcome back.

Hopefully, you spotted that intersection and approximated it to be x equals negative 0.

9, y equals 0.

4 and a second intersection at approximately x equals 0.

9 y equals negative 0.

4.

Well done.

Practise time now.

I'd like you to draw the graphs of both equations and identify the intersections to solve the pair of simultaneous equations.

Pause and do that now.

Question two.

I'd like to fill in the missing coordinate for each pair and then draw each graph to show that there is only one solution for this pair of simultaneous equations.

Pause and give that a go now.

Question three.

I'd like you to draw the graphs of both equations and thus show there is no solution for this pair of simultaneous equations.

Pause and do that now.

Feedback time.

Let's see how we did.

Question one, we are drawing the graph of both equations, identifying the intersections to solve, say for the circle when x equals zero, y equals positive or negative 13.

That pair of coordinates and y equals zero, x is positive or negative 13.

Another pair of coordinate pairs.

We can draw the circle there for the line, when x equals zero, y equals negative seven.

That's that coordinate pair there, when y equals zero, x equals negative seven and we can draw the straight line.

Let's find those intersections now.

We'll have an intersection at negative 12, 5, and an intersection at 5, negative 12.

That gives us the solutions x equals negative 12 y equals five and x equals five, y equals negative 12.

And in this case we have two intersections, therefore two solutions.

Question two, I asked you to fill in the missing coordinate for each pair and draw each graph to show there's any one solution.

So when x equals positive two, y could equal two or y could equal negative two, and when x equals negative two, y could equal negative two or y could equal positive two.

We can plot those four coordinates and you know it's the graph of a circle.

For the straight line when x equals zero, y equals negative four.

When y equals zero, x equals positive four.

That's those two coordinate pairs there.

When we draw those graphs, the circle and the line, you notice something, there's only one intersection at the coordinate two negative two.

There's any one intersection, there's any one solution.

That's because the line is tangent to the circle.

Question three, I asked you to draw the graphs of both equations and thus show there is no solution for this pair.

So the circle when x equals zero, y is positive or negative four, and when y equals zero, x is positive or negative four.

That's those four coordinate pairs for the line when x equals zero, y equals six.

When y equals zero x equals negative eight.

That's those two coordinate pairs.

When you graph your circle and your straight line, you notice something, there's no intersection.

If there are no intersections, there are no solutions for that pair of simultaneous equations.

Sadly, we've reached the end of the lesson now, but we've learned the equation of a circle in the form x squared plus y squared equals r squared can be used to generate coordinate pairs and therefore draw the graph.

The equation of a circle can be used to find the radius of a circle and circles and lines can be graphed to solve pairs of simultaneous equations.

Hope you've enjoyed this lesson as much as I have.

I look forward to seeing you again very soon for more mathematics.

Goodbye for now.