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In order to access this I need to be confident with:

Circles

SubstitutionCoordinates

Gradients

Parallel and perpendicular

Equation of a line

This topic is relevant for:

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.

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

Using Pythagoras’ Theorem it gives the general equation:

\[x^2+y^2=r^2\]

E.g.

This circle has a radius of

\[x^2+y^2=3^2\]

Which simplifies to:

\[x^2+y^2=9\]

In order to recognise the equation of a circle:

**Identify an x^2 term and a y^2 term****Check the radius and use r^2****Identify your final answer**

Get your free circle graph worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free circle graph worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**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:

Identify the correct equation for the graph:

\[ y=x^2+6 \quad \quad \quad y=x^2+12 \quad \quad \quad x^2+y^2=36 \quad \quad \quad x^2+y^2=12\]

**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.

2**Check the radius and use ** r^2

The radius of the circle is

\[r^2=6^2=36\]

3**Identify your final answer**

The correct equation for the graph is:

\[x^2+y^2=36\]

Identify the correct equation for the graph:

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

**Identify an** x^2 ** term and a ** y^2 ** term**

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.

**Check the radius and use ** r^2

The radius of the circle is

\[r^2=10^2=100\]

**Identify your final answer**

The correct equation for the graph is:

\[x^2+y^2=100\]

In order to sketch a circle graph:

**Check the radius****Draw a circle with centre**(0, 0) **Label where the circle crosses the axes**

Sketch x^2+y^2=64

**Check the radius**

The equation of a circle graph with centre

\[x^2+y^2=r^2\]

\[\begin{align*}
r^2 &= 64 \\\\
r &= \sqrt{64}\\\\
r &=8
\end{align*}
\]

**Draw a circle with centre (0, 0) **

**Label where the circle crosses the axes**

\[x^2+y^2=64\]

Sketch x^2+y^2=144

**Check the radius**

The equation of a circle graph with centre

\[x^2+y^2=r^2\]

\[\begin{align*}
r^2 &= 144 \\\\
r &= \sqrt{144}\\\\
r &=12
\end{align*}
\]

**Draw a circle with centre (0, 0) **

**Label where the circle crosses the axes**

\[x^2+y^2=144\]

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

**Find the gradient of the radius from the given point to the centre****Find the gradient of the perpendicular****Find the equation of the lines using the gradient and the given point****Write the equation of the tangent**

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

**Find the gradient of the radius from the given point to the centre**

The gradient of the radius is:

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

**Find the gradient of the perpendicular**

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

\[m_1 \times m_2=-1\]

So the gradient of the tangent is:

**Find the equation of the lines using the gradient and the given point**

The equation of a straight line is of the form:

\[y=mx+c\]

We can substitute the original point for the

The gradient of the tangent is the value of

We can then work out

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

**Write the equation of the tangent**

\[y=-3x+30\]

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

**Find the gradient of the radius from the given point to the centre**

The gradient of the radius is:

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

**Find the gradient of the perpendicular**

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

\[m_1 \times m_2=-1\]

So the gradient of the tangent is:

\[\frac{3}{2}\]

**Find the equation of the lines using the gradient and the given point**

The equation of a straight line is of the form:

\[y=mx+c\]

We can substitute the original point for the

The gradient of the tangent is the value of

We can then work out

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

**Write the equation of the tangent**

\[y=\frac{3}{2}x+\frac{13}{2}\]

r^2**Make sure that you use**

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

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

The gradient of the radius is:

\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

The gradient of the radius is:

\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

1. Identify the equation of this graph:

x^2+y^2=20 \quad \quad x^2+y^2=10 \quad \quad x^2+y^2=100 \quad \quad x^2+y^2=5

**(1 Mark)**

Show answer

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)**

Show answer

Gradient of OP, the radius is 2

\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)**

Show answer

(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)**

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

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