What you need to know

At this level, you are expected to understand and recognise the equation of a circle that is centred at the origin (0,0). That equation looks like

x^2 + y^2 = r^2,

where r is the radius of the circle. For example, consider the equation

x^2 + y^2 = 9

We know that 3^2 = 9, so we now also know that the radius of the circle described by this equation will be 3. As mentioned, all circle graphs you’ll deal with will be centred at the origin, so this particular circle (with radius 3) will look like this.

Because this circle has its centre at the origin, we can see clearly that the radius must be 3 since it crosses both the x-axis and y-axis at the points 3 and -3.

Now we’re going to look at finding the equation of a tangent to a circle. For this we will need to recall a few things.

– The definition of a tangent. (https://mathsmadeeasy.co.uk/gcse-maths-revision/areas-circles-circle-segments-gcse-maths-revision-worksheets/)

– How to calculate the gradient of a line between two points. (https://mathsmadeeasy.co.uk/gcse-maths-revision/gradients-straight-line-graphs-gcse-maths-revision-worksheets/)

– The circle theorem (https://mathsmadeeasy.co.uk/gcse-maths-revision/circle-theorems-gcse-revision-and-worksheets/) which tells us that a tangent is perpendicular to the radius.

– The fact that the gradient of a line is the negative reciprocal of the line perpendicular to it.

– How to form the equation of a straight line. (https://mathsmadeeasy.co.uk/gcse-maths-revision/gradients-straight-line-graphs-gcse-maths-revision-worksheets/)

If any of this feels unfamiliar to you it’s probably worth having another look over it before diving headfirst into this next example.

Example: Find the equation of the tangent to the circle defined by

x^2 + y^2 = 25

at the point (3, 4).

Firstly, we can recognise that because 5^2 = 25, the radius of this circle is 5. With this, we can sketch the graph and draw on what a tangent at the point (3, 4) might look like (it doesn’t have to be perfect, but sometimes this helps us picture what we have to do next).

Now, the first step of this process is to find the gradient of the radius which goes from the centre of the circle to the point (3,4).

This will be useful because of two facts: a tangent is perpendicular to the radius at that point (one of our circle theorems), and if you know the gradient of a line you can obtain the gradient of the perpendicular by taking the negative reciprocal of it. So, the gradient of the line that goes from the origin to (3,4) is

\text{Gradient } = \dfrac{\text{change in } y}{\text{change in }x} = \dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}.

Taking the negative reciprocal of this, we get

\text{Gradient of tangent } = -\dfrac{3}{4}

Now we know the gradient, our straight-line equation must be y = -\frac{3}{4}x + c, where c is the y-intercept that we are yet to determine. However, what we do know is that this tangent passes through the point (3,4), so we can substitute these values of x and y into our straight-line equation and rearrange to find c. We get:

4 = -\dfrac{3}{4} \times 3 + c

4 = -\dfrac{9}{4} + c

c = 4 + \dfrac{9}{4} = \dfrac{25}{4}

Now we’ve found c, we can express our equation of our tangent fully:

y = -\dfrac{3}{4}x + \dfrac{25}{4}

Example Questions

We can see that the radius of this circle extends a distance of 10 away from the centre at (0,0). Therefore, because 10^2 = 100, the equation of the circle is


x^2 + y^2 = 100


First, we need to find the gradient of the line from the centre to (12, 5).


\text{Gradient of radius } = \dfrac{\text{change in } y}{\text{change in }x} = \dfrac{5 - 0}{12 - 0} = \dfrac{5}{12}


Now, by observing that this line we’ve found the gradient of is a radius, and tangents are perpendicular to the radius, we can find the gradient of the tangent by taking the negative reciprocal of the answer we got above.


\text{Gradient of tangent } = -\dfrac{12}{5}


So, we know the straight-line equation for our tangent must be of the form


y = -\dfrac{12}{5}x + c,


where c is the y-intercept which we must determine. To do this, we can substitute the values of x = 12 and y = 5 into the straight-line equation, since we know the line must pass through those coordinates. We get the following.


5 = -\dfrac{12}{5} \times 12 + c


5 = -\dfrac{144}{5} + c


c = 5 + \dfrac{144}{5} = \dfrac{169}{5}


So, our final answer is


y = -\dfrac{12}{5}x + \dfrac{169}{5}


Circle Graphs and Tangents Revision and Worksheets

Graphs and Tangents
Level 8-9
Perpendicular Lines and Tangents
Level 8-9
Graphs and Tangents Exam Questions
Level 8-9
Graphs and Tangents Exam Questions 2
Level 8-9

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