# Circle Graph

Here we will learn about circle graphs, including how to recognise them and how to sketch them. We will also look at solving coordinate geometry questions involving circle graphs.

There are also circle graphs worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is a circle graph?

A circle graph is the graph of an equation which forms a circle.

To do this we have a circle with radius r and centre (0, 0)
Using Pythagoras’ Theorem it gives the general equation:

$x^2+y^2=r^2$

E.g.
This circle has a radius of 3 so has the equation:

$x^2+y^2=3^2$

Which simplifies to:

$x^2+y^2=9$

### What is a circle graph? ## How to recognise the equation of a circle

In order to recognise the equation of a circle:

1. Identify an x^2 term and a y^2 term
2. Check the radius and use r^2

Circle graph is part of our series of lessons to support revision on types of graphs. You may find it helpful to start with the main types of graphs lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

## Circle graph examples

### Example 1: recognise the equation of a circle

Identify the correct equation for the graph:

$y=x^2+6 \quad \quad \quad y=x^2+12 \quad \quad \quad x^2+y^2=36 \quad \quad \quad x^2+y^2=12$

1. Identify an x^2 term and a y^2 term

The equation y=x^2+6 only has an x^2 term and is a quadratic function.

The equation y=x^2+12 only has an x^2 term and is also a quadratic function.

Both the equation x^2+y^2=36 and the equation x^2+y^2=12 have an x^2 term and a y^2 term.

2Check the radius and use r^2

The radius of the circle is 6, so

$r^2=6^2=36$

The correct equation for the graph is:

$x^2+y^2=36$

### Example 2: recognise the equation of a circle

Identify the correct equation for the graph:

$x^2+y^2=20 \quad \quad \quad y=x^3+10 \quad \quad \quad y=x^2+20 \quad \quad \quad x^2+y^2=100$

The equation y=x^2+20 only has an x^2 term and is a quadratic function.

The equation y=x^3+10 only has an x^3 term and is a cubic function.

Both the equation x^2+y^2=20 and the equation x^2+y^2=100 have an x^2 term and a y^2 term.

The radius of the circle is 10, so

$r^2=10^2=100$

The correct equation for the graph is:

$x^2+y^2=100$

## How to sketch a circle graph

In order to sketch a circle graph:

2. Draw a circle with centre (0, 0)
3. Label where the circle crosses the axes

### Explain how to sketch a circle graph ## Sketching a circle graph examples

### Example 3: sketching a circle graph

Sketch x^2+y^2=64

The equation of a circle graph with centre (0, 0) is of the form:

$x^2+y^2=r^2$

\begin{align*} r^2 &= 64 \\\\ r &= \sqrt{64}\\\\ r &=8 \end{align*}
$x^2+y^2=64$

### Example 4: sketching a circle graph

Sketch x^2+y^2=144

The equation of a circle graph with centre (0, 0) is of the form:

$x^2+y^2=r^2$

\begin{align*} r^2 &= 144 \\\\ r &= \sqrt{144}\\\\ r &=12 \end{align*}
$x^2+y^2=144$

## How to find the equation of a tangent to a circle

In order to find the equation of a tangent to a circle:

1. Find the gradient of the radius from the given point to the centre
2. Find the gradient of the perpendicular
3. Find the equation of the lines using the gradient and the given point
4. Write the equation of the tangent

## Finding the equation of a tangent to a circle examples

### Example 5: finding the equation of a tangent to a circle

Find the equation of the tangent to the circle x^2+y^2=90 at the point (9,3)

$\frac{Change\hspace{2mm}in\hspace{2mm}y}{Change\hspace{2mm}in\hspace{2mm}x}=\frac{3}{9}=\frac{1}{3}$

The radius and the tangent meet at right angles; this is a circle theorem.
Therefore the radius and the tangent are perpendicular.

The product of perpendicular gradients is −1.

$m_1 \times m_2=-1$

So the gradient of the tangent is:

−3

The equation of a straight line is of the form:

$y=mx+c$

We can substitute the original point for the x and y values.
The gradient of the tangent is the value of m
We can then work out c.

\begin{align*} y &= mx+c \\\\ 3 &= -3 \times 9 +c\\\\ 3 &= -27 +c\\\\ 30 &= c \end{align*}
$y=-3x+30$

### Example 6: finding the equation of a tangent to a circle

Find the equation of the tangent to the circle x^2+y^2=13 at the point (-3,2)

$\frac{Change\hspace{2mm}in\hspace{2mm}y}{Change\hspace{2mm}in\hspace{2mm}x}=\frac{2}{-3}=-\frac{2}{3}$

The radius and the tangent meet at right angles; this is a circle theorem.
Therefore the radius and the tangent are perpendicular.

The product of perpendicular gradients is −1.

$m_1 \times m_2=-1$

So the gradient of the tangent is:

$\frac{3}{2}$

The equation of a straight line is of the form:

$y=mx+c$

We can substitute the original point for the x and y values.
The gradient of the tangent is the value of m
We can then work out c.

\begin{align*} y &= mx+c \\\\ 2 &= \frac{3}{2} \times -3 +c\\\\ 2 &= \frac{-9}{2} +c\\\\ \frac{13}{2} &= c \end{align*}

$y=\frac{3}{2}x+\frac{13}{2}$

### Common misconceptions

• Make sure that you use r^2

The equation of the circle requires the radius NOT the diameter.

Remember the radius is half the diameter.

• The radius is always a positive number

Since the radius is a length it is always positive.

E.g.

The radius of this circle is 12 and r^2=144 . The equation of this circle is x^2+y^2=144

### Practice circle graph questions

1. What is the equation for this graph: x^2+y^2=125 x^2+y^2=10 x^2+y^2=5 x^2+y^2=25 The equation of a circle graph with centre (0,0) is of the form:

x^2+y^2=r^2

The radius of the circle is 5 , so

r^2=5^2=25

2. What is the equation for this graph: x^2+y^2=49 x^2+y^2=-49 x^2+y^2=14 x^2+y^2=-14 The equation of a circle graph with centre (0,0) is of the form:

x^2+y^2=r^2

The radius of the circle is 7 , so

r^2=7^2=49

3. Sketch:

x^2+y^2=16        The equation of a circle graph with the centre of the circle (0,0) is of the form:

x^2+y^2=r^2

\begin{aligned} r^2 &= 16 \\ r &= \sqrt{16}\\ r &=4 \end{aligned}

So we draw a circle with centre (0,0) radius of the circle is 4.

4. Sketch:

x^2+y^2=81        The equation of a circle graph with centre (0,0) is of the form:

x^2+y^2=r^2

\begin{aligned} r^2 &= 81 \\ r &= \sqrt{81}\\ r &=9 \end{aligned}

So we draw a circle with centre (0,0) radius of the circle is 4.

5. Find the equation of the tangent to the circle x^2+y^2=80 at the point (8,4) y=-\frac{1}{2}x+20 y=2x+20 y=-2x+32 y=-2x+20 \frac{change\hspace{1mm}in\hspace{1mm}y}{change\hspace{1mm}in\hspace{1mm}x}=\frac{4}{8}=\frac{1}{2}

The gradient of the tangent will be: -2

\begin{aligned} y &= mx+c \\ 4 &= -2 \times 8 +c\\ 4 &= -16 +c\\ 20 &= c \end{aligned}

So the equation would be:

y=-2x+20

6. Find the equation of the tangent to the circle x^2+y^2=20 at the point (-2, 4) y=\frac{1}{2}x+5 y=2x+5 y=2x+4 y=-\frac{1}{2}x+4 \frac{change\hspace{1mm}in\hspace{1mm}y}{change\hspace{1mm}in\hspace{1mm}x}=\frac{4}{-2}=-2

The gradient of the tangent will be: \frac{1}{2}

\begin{aligned} y &= mx+c \\ 4 &= \frac{1}{2} \times -2 +c\\ 4 &= -1 +c\\ 5 &= c \end{aligned}

So the equation would be:

y=\frac{1}{2}x+5

### Circle graph GCSE questions

1.   Identify the equation of this graph: (1 Mark)

The equation of a circle graph with centre (0,0) is of the form:

x^2+y^2=r^2

The radius of the circle is 10 , so

r^2=10^2=100

The correct equation is:

x^2+y^2=100

(1)

2.   A point P (2,4) lies on the circle with equation:

x^2+y^2=20

Find the equation of the tangent to the circle at the point P.

(3 Marks)

\frac{change\hspace{1mm}in \hspace{1mm}y}{change\hspace{1mm}in\hspace{1mm}x}=\frac{4}{2}=2

(1)

The gradient of the tangent is -\frac{1}{2}

(1)

\begin{aligned} y &= mx+c \\\\ 4 &= -\frac{1}{2} \times 2 +c\\\\ 4 &= -1 +c\\\\ 5 &= c \end{aligned}

The equation of the tangent is:

y=-\frac{1}{2}x+5

(1)

3.   (a)  On the grid, draw the graph of:

x^2+y^2=16 (b) Hence find estimates for the solutions of simultaneous equations:

x^2+y^2=16

x+y=3

(3 Marks)

(a) for a circle with centre (0.0)

(1)

for the correct circle with radius 4

(1)

(b) x=3.9 and y= -0.9

x= -3.9 and y=0.9

for drawing the line graph x+y=3

(1)

for one of the correct pair of x and y solutions OR both x -values OR both y -values

(1)

for both pairs of correct solutions

(1)

## Learning checklist

You have now learned how to:

• recognise the equation of a circle with centre at the origin
• use the equation of a circle with centre at the origin
• find the equation of a tangent to a circle at a given point

## Still stuck?

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