Lesson video

Mr. Robson here.

Thanks for joining me today.

We're going to learn how to find the equation of the tangent to a circle.

It's a pretty awesome bit of maths, so let's get on with it.

A learning outcome is I'll be able to find the equation of the tangent to a circle at any given point.

Keywords that we're gonna see today.

Gradient's one of them.

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.

Radius is another important word.

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

Finally, tangent.

A tangent of a circle is a line that intersects the circle exactly once.

Look out for those keywords throughout our learning today.

Two parts to our learning.

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

We can deduce a lot from the equation of a circle.

X minus five squared plus y minus three squared equals two squared.

What do we know about this circle? Well, we would know that the centre is at five, three, and we know that it's got a radius of two.

How do we know this? Because the general form of the equation of a circle.

Circles in the form, x minus a squared plus y minus b squared equal r squared, have a centre at a, b and radius r.

That generalisation is really important, and you're gonna lean on it a lot throughout this lesson, so you'll want to write it down and keep it handy.

Pause and do that now.

Using that generalisation is going to be really important throughout our lesson today, so let's check you can use it.

You're going to match the equations to the centre of their circle.

There's four equations on the left hand side of the screen.

They all form circles.

Four coordinate pairs on the right hand side are the centres of those circles, match them up.

Pause and do this now.

Welcome back, let's see how we did.

You would've noticed that they all have r squared as five squared on the right hand side of each equation, so it's all about what's on the left hand side of the equations.

In the case of x minus eight squared plus y minus two squared, that'll give us a centre of eight, two.

For x plus eight squared plus y plus two squared, that'll be at the centre, -8, -2.

For the third one, we should have a centre -7, positive six, and for the fourth, -7, -6.

Remember, equations in the form, x minus a squared plus y minus b squared equal r squared, have a centre at coordinate pair a, b.

Izzy is considering the graph of this circle, x minus nine squared plus y plus four squared equals one.

Izzy says, "The centre of this circle "will be at -9, four." I can see why Izzy has said that.

There's an x next to a -9, and there's a y next to a positive four.

So it would make sense to think that the centre of this circle is at -9, four, but we know better, don't we? What I'd like you to do is write a sentence to explain to Izzy why this is wrong.

Pause and do that now.

Welcome back.

You might have written, "Equations in the form, "x minus a squared plus y minus b squared equal r squared, "have a centre at a, b." Try to compare the equation to this form.

What do we mean by that? It's really useful when looking at the equation of a circle, if you write the general form next to your equation.

From here, we can see we have an a value of nine and a b value of -4, so that tells us the centre must be at nine, -4, not what Izzy had.

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

For example, the equation of a circle is x minus 15 squared plus y plus 16 squared equal 169.

A point in the circumference is three, -21.

Find the equation of the radius.

Well firstly, we can see from the equation that this circle has a centre at 15, -16, also a radius of 13.

A sketch of the problem will support our understanding.

Frequently in mathematics, sketching a problem out helps us to see what's going on.

Our sketch might look like that.

Centre of the circle at 15, -16, the point in the circumference at three, -21, and roughly a radius of 13, which we know from that centre point means our circle will not intercept the x or y axis.

From our sketch, we can draw a radius and identify we've got a change in x of positive 12 and a change in y of positive five.

From here, we can find the equation of the radius.

It's going to be a linear equation of the form, y equals mx plus c with a gradient of 5/12.

We know it's going to take the form, y equals 5/12x plus c.

We can substitute to find that constant.

We know the centre is at x equals 15, y equals -16.

Substitute those in.

We find the value of the constant to be -89/4, so the equation of the radius is y equals 5/12x minus 89/4.

But it's really sensible at this point to check our equation using our second coordinate pair.

If we substitute three, -21 into our equation, we know that we're right.

We don't have to use a sketch.

Once you have two coordinate pairs, you can use the formula for gradient.

There's the formula for gradient.

Substituting in our two coordinate pairs gives us a gradient of 5/12, and from there, we perform the same substitution into y equals mx plus c to find the equation of our radius.

Didn't need the sketch, but it sure is useful.

It allows us to see what's going on.

Sometimes the equation of the radius needs to be given in a certain form.

For example, find the equation of the radius in the form, ay plus bx equals c where a, b, and c are integers.

We might not want it in the form we just found, y equals 5/12x minus 89/4.

We want it in this form, ay plus bx equals c.

So, if we multiply everything in that equation through by 12, we get 12y equals 5x minus 267.

From there, we can rearrange that to read, 12y minus 5x equals -267, or you might see it written as 267 equals 5x minus 12y.

Quick check you've got this.

The equation of a circle is x plus 30 squared plus y minus 20 squared equals 625.

A point in the circumference is -37, 44.

I'd like you to sketch that circle.

Pause, give this a go now.

Welcome back, let's see how we did.

Information that we need before performing that sketch.

We need to know that the centre is at -30, 20, and identify a radius of 25.

From there, we can sketch this.

You'll notice that the circle does intersect the x axis because we've got a radius of 25, yet the height of the y coordinate is only 20.

From here, I'd like you to use your sketch to find the equation of the radius.

Pause and do that now.

Welcome back, let's see how we did.

Lots of ways you could tackle this.

You might draw a right triangle and identify for a change of positive seven and x.

We've got a change of -24 in y, that'll give us a gradient of -24/7.

So in the form, y equals mx plus c, it's gonna be y equals -24/7x plus c.

We can substitute in one of our known coordinates.

I've gone for the centre, x equals -30, y equals 20.

From there, we can find our constant, -580/7, so the equation of our radius must be y equals -24/7x minus 580/7.

I do hope you didn't stop there and you checked your work.

If we substitute in the other known coordinate pair, <v ->37, 44, we find they satisfy the equation, those values,</v> therefore, the equation of our radius must be correct.

One more check of this skill.

I love using a sketch.

They help me to see what's going on in a problem, but we don't need a sketch.

It can be done without it, let's practise this.

The equation of a circle is x plus 30 squared plus y minus 20 squared equals 625.

Another point on the circumference is -6, 27.

Find the equation of this radius without using a sketch.

Hint for you, you'll want to use the formula for finding the gradient.

Pause, see if you can find the equation of this radius without using a sketch.

Welcome back.

Hopefully, you identified first, there's a centre at -30, 20, for this circle.

When we substitute that into our formula for gradient, find a gradient of 7/24.

Y equal 7/24 x plus c, let's find the constant by substituting in the coordinate pair of our centre.

Our constant is 115/4, so the equation of our radius must be y equals 7/24x plus 115/4.

Of course, we're going to check this.

We'll use the other coordinate pair that we know, -6, 27.

When we substitute those in, they satisfy the equation, so we know that we're right.

What if we want it in a different form? What if we we're asked to find the equation of the radius in the form ay plus bx equals c, where a, b and c are integers? We know the equation of the radius is y equals 7/24x plus 115/4.

Can you rearrange this so it's in the correct form? Pause and do this now.

Welcome back.

Hopefully, you multiplied every term by 24.

We get 24y equals 7x plus 690.

From there, we can rearrange, and we'll get 24y minus 7x equals 690.

Practise time now.

Question one, the equation of a circle is x minus 58 squared plus y plus 17 squared equals 37 squared.

A point on the circumference is 93, -29.

I'd like you to find the equation of the radius connecting this point to the centre.

For question two, slightly different this question, but very exciting.

Question two, the equation of a circle is x minus 10 squared plus y minus 25 squared equals 225.

A point in the circumference is 22, 34.

What I'd like you to do for this question is show that the coordinate pair 14, 28, is on the radius connecting the point on the circumference to the centre.

Pause, try these problems now.

Welcome back, let's see how we did.

For question one, we should have started identifying the centre of the circle at 58, -17.

We know we've got a radius of 37 as well.

Sketch would look like, so sketch helps us to see what's going on in this problem.

This sketch shows us that we've got a gradient of -12/35.

We can use the form, y equals mx plus c, substitute in our known centre.

We'll find the constant to be 101/35.

So the equation of our radius must be y equals -12/35x plus 101/35.

And you know what we're gonna do now, we're gonna check by substituting in our second known coordinate pair of 93, -29.

If that pair of coordinates satisfies the equation, we know our equation is correct.

You don't have to draw a sketch.

You could have identified the centre, substituted the two known coordinate pairs into our formula for gradient, and used y equals mx plus c, substitute to find that constant, you'd have got the exact same equation for our radius.

Question two, we were showing that the coordinate pair 14, 28 is on the radius.

First, we need to find the equation of that radius.

So, centre of the circle, 10, 25, the gradient connecting 10, 25 to 22, 34 is 3/4.

Y equals 3/4x plus c.

Substitute in the coordinate pair for our centre, x equals 10, y equals 25, we find the constant to be 35/2.

So the equation of our radius must be y equals 3/4x plus 35/2.

We'll check that, of course, by substituting in x equals 22, y equals 34.

When we know that that pair of values satisfy the equation, we know that that is indeed the equation of our radius.

Are we finished? Absolutely not.

We need to show that 14, 28 is on the radius, so we do that by substituting in x equals 14, y equals 28, into the equation of our radius.

When they satisfy the equation, we know we're right, but don't stop there.

You want to write a supporting sentence, which reads something like this.

"The values 14, 28 satisfy the equation, "therefore, this point is on the radius." Onto the second part of our learning now.

We've got a good understanding of the equation of the radius, so we're ready to think about the equation of the tangent.

The equation of a circle is x minus nine squared plus y plus six squared equals 13 squared.

A point on the circumference is 14, -18.

What can we work out from this information? Pause, see if you can make a list of everything you know from this information.

Welcome back.

I wonder what was on your list.

I hope you wrote the radius of the circle, the centre of the circle, the gradient of the radius, and the equation of the radius.

Absolutely, we can work those things out.

You can see them all in our sketch here.

What else could we find? How about some information about the tangent at that point? Well, if we know the gradient of the radius, we can find out the gradient of the tangent, and if we know the gradient of the tangent and a point on that line, we could find the equation of the tangent.

A crucial thing in working out the equation of our tangent is understanding the relationship between the gradient of the radius and the gradient of the tangent.

We're gonna recall some knowledge from geometry here.

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

The gradients of the tangent and radius must therefore be negative reciprocals.

There's two important points to remember there.

You might wanna pause now and write them down.

So we can find the gradient of this radius.

We know that it will be -12/5.

The gradient of the tangent.

We know that when we multiply the gradient of the radius by the gradient of the tangent, we'll get -1, because the two gradients will be negative reciprocals for one another.

So if we want to find the gradient of the tangent, we do -1 divided by the gradient of the radius.

In this case, that's -1 divided by -12/5, giving us 5/12.

The gradient of our tangent is 5/12.

This is where I enjoy having my sketch handy.

Does it look like a gradient of slightly less than a half? Absolutely.

Quick check you've got this.

The equation of a circle is x plus nine squared plus y minus six squared equals 25.

A point in the circumference is -6, 10.

I'd like you to find the gradient of the tangent to the circle at this point.

Pause and see if you can work that out.

Welcome back, let's see how we did.

Hopefully, you identify the centre of the circle is -9, 6.

A sketch is gonna help us here.

That sketch will help us find the gradient of the radius, which is going to be 4/3.

I'm now gonna sketch my tangent.

The gradient of a tangent must be <v ->1 divided by the gradient of the radius.

</v> In this case, -1 divided by 4/3.

The gradient of our tangent is -3/4.

Does it look like a gradient of -3/4? Absolutely.

The equation of the circle, x plus nine squared plus y minus six squared equals 25, a point on the circumference -6, 10, we know the gradient of the tangent is -3/4.

What can we work out from here? We've got the gradient.

We know a point on the line.

We can find the equation of the tangent.

It's going to be y equals -3/4x plus c.

We know the coordinate pair, -6, 10, is on this line.

The constant, therefore, is 11/2.

That's the equation of our tangent, y equals -3/4x plus 11/2.

In this case, we've got no second coordinate to check the answer, but our sketch supports it.

We've got a negative gradient, slightly less steep than -1 and the y intercept where we expect to see it.

What do I mean by that? Well, the equation y equals -3/4x plus 11/2, that's gonna have a y intercept at zero, 11/2.

11/2 is 5.

5 as a decimal.

Does our y intercept on our sketch look like it might be at zero, 5.

5? Absolutely.

I like this answer, I think we're right.

Quick check you can do that.

The equation of a circle is x plus 150 squared plus y plus 40 squared equals 100 squared.

A point in the circumference is -122, -136.

I'd like you to find the equation of the tangent to the circle at this point.

Pause and try this now.

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

Hopefully, you started by finding the centre of this circle, - 150, -40.

A sketch of this circle would look like that, as our centre and our coordinate on the circumference, <v ->122, -136.

</v> The gradient at the radius at that point, is gonna be -24/7.

If the gradient of our radius is -24/7, then the gradient of our tangent must be 7/24.

Does our sketch confirm that that tangent has a gradient of 7/24? Looks good to me.

From there, we can substitute in to y equals mx plus c, our known coordinate pair, -122, -136.

We find a constant of -1205/12, so the equation of our tangent must be y equals 7/24x minus 1205/12.

Or you might write that in the form, 24y equals 7x minus 2410.

Practise time now.

Question one, the equation of a circle is x plus 156 squared plus y minus 87 squared equals 1156, point on the circumference, -140, 117.

I'd like you to find the equation at the tangent to the circle at this point.

Question two is a little more complicated, but it's a lovely problem.

The equation of a circle is x minus 35 squared plus y minus 47 squared equals 25.

Two points in the circumference are 32, 51, and 39, 44.

What I'd like you to do for question two is find the coordinates of the point of intersection where the tangents to the circle at those two points meet.

My top hint for question two is a sketch is really going to help you here.

Pause, try both those problems now.

Welcome back.

Feedback time, let's see how we did.

Hopefully, you started by finding the centre of this circle in question one, -156, 87.

You didn't have to do a sketch, but they really are helpful.

We can find the gradient of the radius, which is 15/8, if the gradient of the radius is 15/8, then the gradients of our tangent is -8/15.

So the equation of our tangent must be y equals -8/15x plus c.

Let's find that constant by substituting in the known coordinate.

Our constant must be 127/3, therefore, the equation of our tangent is y equals -8/15x plus 127/3.

My sketch confirms that that's pretty good as an answer.

You might not have left the answer in that form.

You might have written it as 15y equals 635 minus 8x.

For question two, there's a lot going on here, and I said a sketch helped.

When we sketch it out, we see the centre of the circle of 35, 47, and the two points in the circumference.

We draw those tangents, we can see where they meet.

We need to know the equation of each tangent.

The first tangent is y equals 3/4x plus 27, and our second tangent, y equals 4/3x minus eight.

Where do those two tangents meet? Well, they will meet when 3/4x plus 27 is equal to 4/3x minus eight.

Rearrangement of that would leave you with 7/12x equals 35, so x equals 60, but you wouldn't leave your answer like that.

We're asked for the coordinates plural.

We've got the x coordinate, we need the y one.

So we'll substitute the x coordinates into the equation of the tangent, and we'll find that y equals 72.

So we can say the intersection of those two tangents is at coordinate 60, 72.

Sadly, that's the end of the lesson now.

Hope you've enjoyed it as much as I have.

What have we learned? If we know the equation of a circle and a point on the circumference, we can find the gradient of the radius, and therefore, the gradient of the tangent because it will be the negative reciprocal.

If we know the gradient of the tangent and the point it touches the circle, we can find the equation of the tangent.

Thank you for all your hard work today.

I look forward to seeing you again soon for more maths.