**Circle Graphs and Tangents **

**Circle graphs** are another type of graph you need to know about. Questions involving circle graphs are some of the hardest on the course.

You need to be able to plot them as well as calculate the equation of **tangents** to them.

Make sure you are happy with the following topics

**What are Circle Graphs?**

Circle graphs for a circle, have the general equation

\textcolor{red}{x}^2 + \textcolor{limegreen}{y}^2 = \textcolor{blue}{r}^2

e.g:

\textcolor{red}{x}^2 + \textcolor{limegreen}{y}^2 = \textcolor{blue}{4}^2

which is the same as

\textcolor{red}{x}^2 + \textcolor{limegreen}{y}^2 = \textcolor{blue}{16}

**Remember:** r is the radius of the circle formed around centre (0,0)

**Finding the Equation of a Tangent**

This is the most typical exam question you will face, if you learn the steps it really isn’t as tricky as it seems.

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

x^2 + y^2 = \textcolor{red}{25}

at the point (3, 4), shown on the axes below.

**Step 1:** Find the gradient of the radius.

Firstly, we can recognise that because \textcolor{red}{5^2} = \textcolor{red}{25}, the radius of this circle is \textcolor{red}{5}.

We need to find the gradient of the radius which goes from the centre of the circle to the point (3,4).

The **tangent** is **perpendicular** to the radius at that point (one of our circle theorems), meaning 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

\begin{aligned} \text{Gradient } &= \dfrac{\text{change in } y}{\text{change in }x} \\ &= \dfrac{4 - 0}{3 - 0} \\ &= \dfrac{4}{3} \end{aligned}

**Step 2:** Find the gradient of the Tangent.

Taking the **negative reciprocal** of this, we get

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

**Step 3:** Complete the rest of the equation.

Now we know the gradient, our straight-line equation must be y = \textcolor{blue}{-\frac{3}{4}}x + c, where c is the y-intercept that we are yet to determine.

We know 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 = \textcolor{blue}{-\dfrac{3}{4}} \times 3 + c

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

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

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

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

## Make up for lost time with the GCSE Maths Catch Up Course

### Example Questions

**Question 1:** State the equation of the circle pictured below.

**[2 marks]**

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

**Question 2:** Find the equation of the tangent to the circle below at the point marked with a cross.

**[4 marks]**

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 is a radius, and that 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}

**Question 3: **Find the equation of the tangent to x^2 + y^2 = 113 at the point (-8,-7).

**[4 marks]**

First, we need to find the gradient of the line from the centre to (-8, -7).

\text{Gradient of radius } = \dfrac{\text{change in } y}{\text{change in }x} = \dfrac{-7 - 0}{-8 - 0} = \dfrac{7}{8}

Now, by observing that this line is a radius, and that 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{8}{7}

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

y = -\dfrac{8}{7}x + c,

where c is the y-intercept which we must determine. To do this, we can substitute the values of x = -8 and y = -7 into the straight-line equation, since we know the line must pass through those coordinates. We get the following.

-7 =-\dfrac{8}{7} \times -8 + c

-7 = \dfrac{64}{7} + c

c=-\dfrac{113}{7}

So, our final answer is

y = -\dfrac{8}{7}x -\dfrac{113}{7}

### Worksheets and Exam Questions

#### (NEW) Circle Graphs and Tangents Exam Style Questions - MME

Level 8-9 New Official MME### Drill Questions

#### Graphs and Tangents

#### Graphs and Tangents Exam Questions

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