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Here we will learn about the equation of a tangent, including finding the equation of a tangent of a circle.

There is also an equation of a tangent worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The** equation of a tangent** line is the equation of the straight line touching the circumference of the circle at only one point, known as the tangent. A line is only a tangent if there is exactly one point of contact between the straight line and the circle.

To find the equation of a tangent, we first need to be able to find the gradient of the radius of the circle – we use the gradient formula for finding the gradient of a line segment joining two points,

m=\cfrac{y_{2}-y_{1}}{x_{2}-x}to find the gradient of the radius (m_{1}).

**Step-by-step guide:** Gradient of a line

We know from work on circle theorems that the angle between the radius of a circle and the tangent line at a given point is 90 degrees – these two lines are **perpendicular**.

We then use knowledge of circle theorems and perpendicular lines to find the gradient of the tangent (m_{2}) as follows,

- the tangent and radius are
**perpendicular**, - the product of perpendicular gradients is -1

(i.e. m_{1} m_{2}=-1 ), - rearranging this gives the gradient of the tangent m_{2}=\cfrac{-1}{m_{1}}.

Alternatively, it may help to think of the gradient of the tangent as the negative reciprocal of the gradient of the radius – so find the reciprocal, then multiply by -1, which has the effect of changing the sign.

For example, if the gradient of the radius is the fraction \cfrac{4}{3}, we would find the reciprocal, which is \cfrac{3}{4}, then multiply by -1 to give the gradient of the tangent as -\cfrac{3}{4}.

**Step-by-step guide:** Parallel and perpendicular lines

Finally we find the y -intercept of the tangent by substituting our known values into the equation y=mx+c and solving to find c. This completes the full equation of the tangent to the circle.

When working with **circle equations **at GCSE, you are only expected to deal with examples where the centre of the circle is at the origin (0,0). These ideas are extended to any circle with a non-zero centre in A Level maths.

However, when finding the equation of a tangent, you may be expected to work with circles with a non-zero centre – this is because the method used is identical whether the circle is centred at the origin or elsewhere on the coordinate grid. You can see this in example 5 below.

In order to find the equation of the tangent to a circle at a given point:

**Find the gradient of the radius at that point.****Find the gradient of the tangent at that point.****Find the**\textbf{y}**-intercept and hence the equation of the tangent.**

Get your free equation of tangent to a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free equation of tangent to a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Equation of tangent** is part of our series of lessons to support revision on **parts of a circle** and **circles, sectors and arcs**. You may find it helpful to start with the main circles, sectors and arcs 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:

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 at that point.**

The radius is the line joining the centre of the circle (0,0) and the known point on the circumference (9,3).

To find the gradient of the radius, calculate m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\cfrac{3-0}{9-0}=\cfrac{3}{9}=\cfrac{1}{3}2**Find the gradient of the tangent at that point.**

The gradient of the radius =\cfrac{1}{3}.

The reciprocal of =\cfrac{1}{3} is 3.

Multiply this by -1 (change the sign) to get the gradient of the tangent, gradient =-3.

3**Find the ** \textbf{y} **-intercept and hence the equation of the tangent.**

We now have the gradient of the tangent line, -3, so we know the equation of the line is y=-3x+c.

We use the known point on that line (9,3) to find c.

We substitute the values x=9 and y=3 into y=-3x+c and solve for c.

\begin{aligned} 3 &=-3(9)+c \\\\ 3 &=-27+c \\\\ 30 &=c \end{aligned}Hence the equation of the tangent is y=-3x+30, or, better, y=30-3x.

Find the equation of the tangent to the circle x^{2}+y^{2}=80 at the point (-4,8).

**Find the gradient of the radius at that point.**

The radius is the line joining the centre of the circle (0,0) and the known point on the circumference (-4,8).

To find the gradient of the radius, calculate m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\cfrac{0-8}{0–4}=\cfrac{-8}{4}=-2

**Find the gradient of the tangent at that point.**

The gradient of the radius =-2.

The reciprocal of -2 is -\cfrac{1}{2}.

Multiply this by -1 (change the sign) to get the gradient of the tangent, gradient =\cfrac{1}{2}.

**Find the ** \textbf{y} **-intercept and hence the equation of the tangent.**

We now have the gradient of the tangent line, \cfrac{1}{2}, so we know the equation of the line is y= \cfrac{1}{2} x+c.

We use the known point on that line (-4,8) to find c.

We substitute the values x=-4 and y=8 into y= \cfrac{1}{2}x+c and solve for c.

\begin{aligned} 8 &=\cfrac{1}{2}(-4)+c \\\\ 8 &=-2+c \\\\ 10 &=c \end{aligned}

Hence the equation of the tangent is y=\cfrac{1}{2} x+10.

Find the equation of the tangent to the circle x^{2}+y^{2}=25 at the point (4,3).

**Find the gradient of the radius at that point.**

The radius is the line joining the centre of the circle (0,0) and the known point on the circumference (4,3).

To find the gradient of the radius, calculate m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\cfrac{3-0}{4-0}=\cfrac{3}{4}

**Find the gradient of the tangent at that point.**

The gradient of the radius =\cfrac{3}{4}.

The reciprocal of \cfrac{3}{4} is \cfrac{4}{3}.

Multiply this by -1 (change the sign) to get the gradient of the tangent, gradient =-\cfrac{4}{3}.

**Find the ** \textbf{y} **-intercept and hence the equation of the tangent.**

We now have the gradient of the tangent line, =-\cfrac{4}{3}, so we know the equation of the line is y=-\cfrac{4}{3} x+c.

We use the known point on that line (4,3) to find c.

We substitute the values x=4 and y=3 into y= -\cfrac{4}{3}x+c and solve for c.

\begin{aligned} &3=-\cfrac{4}{3}(4)+c \\\\ &3=-\cfrac{16}{3}+c \end{aligned}

It is useful to write 3=\cfrac{9}{3} to be able to rearrange the next step.

\begin{aligned} \cfrac{9}{3} &=-\cfrac{16}{3}+C \\\\ \cfrac{25}{3} &=C \end{aligned}

Hence the equation of the tangent in the form y=mx+c (sometimes called slope intercept form) is y=-\cfrac{4}{3} x+\cfrac{25}{3}.

In cases like this with fractional coefficients, it is also acceptable to multiply by 3 and rearrange to write the final answer as follows,

\begin{aligned} y &=-\cfrac{4}{3} x+\cfrac{25}{3} \\\\ 3 y &=-4 x+25 \\\\ 3 y+4 x &=25 \end{aligned}

An answer in the standard form ax+by=d or ax+by+d=0 is usually much easier to work with if you’re expected to do more with the equation, such as find a point of intersection (see example 4).

The diagram shows the circle x^{2}+y^{2}=225.

The tangent to the circle at the point (9,12) intersects the x axis at point A.

Find the x coordinate of point A.

**Find the gradient of the radius at that point.**

The radius is the line joining the centre of the circle (0,0) and the known point on the circumference (9,12).

To find the gradient of the radius, calculate m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\cfrac{12-0}{9-0}=\cfrac{4}{3}

**Find the gradient of the tangent at that point.**

The gradient of the radius =\cfrac{4}{3}.

The reciprocal of \cfrac{4}{3} is \cfrac{3}{4}.

Multiply this by -1 (change the sign) to get the gradient of the tangent =-\cfrac{3}{4}.

**Find the ** \textbf{y} **-intercept and hence the equation of the tangent.**

We now have the gradient of the tangent line, -\cfrac{3}{4}, so we know the equation of the line is y=-\cfrac{3}{4} x+c.

We use the known point on that line (9,12) to find c.

We substitute the values x=9 and y=12 into y= -\cfrac{3}{4}x+c and solve for c.

\begin{aligned} &12=-\cfrac{3}{4}(9)+c \\\\ &12=-\cfrac{27}{4}+c \end{aligned}

It is useful to write 12=\cfrac{48}{4} to be able to rearrange the next step.

\begin{aligned} &\cfrac{48}{4}=-\cfrac{27}{4}+C \\\\ &\cfrac{75}{4}=C \end{aligned}

Hence the equation of the tangent is y=-\cfrac{3}{4} x+\cfrac{75}{4}.

Because we have a further step to complete, we can multiply by 4 and rearrange to write the equation in the form,

4y+3x=75.

**Substitute ** \bf{y=0} ** to find the ** \textbf{x} ** value of the point of intersection.**

The tangent intersects the x axis at point A, we know that y=0 because this is the value of y all along the x axis.

We substitute y=0 into the equation of the tangent.

\begin{array}{r} 4 y+3 x=75 \\\\ 4(0)+3 x=75 \\\\ 3 x=75 \\\\ x=25 \end{array}

Hence the x coordinate of point A is 25.

The centre of a circle has the coordinates (-1,2). The point (3,5) lies on the circumference of the circle. Find an equation of the tangent to the circle at this point.

**Find the gradient of the radius at that point.**

The radius is the line joining the centre of the circle (-1,2) and the known point on the circumference (3,5).

To find the gradient of the radius, calculate m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

m=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\cfrac{5-2}{3-(-1)}=\cfrac{3}{4}

**Find the gradient of the tangent at that point.**

The gradient of the radius =\cfrac{3}{4}.

The reciprocal of =\cfrac{3}{4} is \cfrac{4}{3}.

Multiply this by -1 (change the sign) to get the gradient of the tangent, gradient =-\cfrac{4}{3}.

**Find the ** \textbf{y} **-intercept and hence the equation of the tangent.**

We now have the gradient of the tangent line, -\cfrac{4}{3}, so we know the equation of the line is y=-\cfrac{4}{3}x+c.

We use the known point on that line (3,5) to find c.

We substitute the values x=3 and y=5 into y=-\cfrac{4}{3} x+c and solve for c.

\begin{aligned} 5 &=-\cfrac{4}{3}(3)+c \\\\ 5 &=-4+c \\\\ 9 &=c \end{aligned}

Hence the equation of the tangent is y=-\cfrac{4}{3}x+9.

As this equation contains a fraction, it is better to rewrite in the form ax+by=d, so we multiply by 3 and rearrange to give

\begin{aligned} 3 y &=-4 x+27 \\\\ 4 x+3 y &=27 \end{aligned}

Hence the equation of the tangent to the circle at the given point is 4x+3y=27.

Sometimes you might be asked to prove statements about equations of tangents, such as proving that a given line is tangent to a given circle.

For example, you might be asked to prove algebraically that the straight line 2x+y=20 is a tangent to the circle with equation x^{2}+y^{2}=80.

Notice that we have two equations with two unknowns, so we are solving simultaneous equations. As the equation of a circle involves squared terms, we are in fact solving a pair of simultaneous equations where one is linear and the other one quadratic.

**Step-by-step guide:** Quadratic simultaneous equations

In this case, we rearrange the linear equation 2x+y=20 to give us a statement we can substitute into the circle equation, in order to eliminate one variable and form a quadratic equation that we can solve using standard methods.

We know that the tangent to a circle only touches the circle at one point; this means that if the line given is a tangent, we should only obtain one solution to the quadratic equation. As we are writing a proof, the final line needs to be a statement that we have shown this to be true.

In order to prove that a line is a tangent to a circle:

**Rearrange the linear equation and substitute into the equation of the circle to get a quadratic.****Simplify and solve the quadratic equation.****Make a statement about the solution.**

Prove algebraically that the straight line 2x+y=20 is a tangent to the circle with equation x^{2}+y^{2}=80.

**Rearrange the linear equation and substitute into the equation of the circle to get a quadratic.**

We can rearrange 2x+y=20 to make either x or y the subject – it doesn’t matter which we choose. Here we will rearrange to make y the subject because this requires fewer steps.

\begin{aligned} 2x+y&=20 \\ y&=20-2x \end{aligned}

Then substitute this expression for y into the equation of the circle.

\begin{aligned} x^{2}+y^{2}&=80 \\ x^{2}+(20-2x)^{2}&=80 \end{aligned}

**Simplify and solve the quadratic equation.**

We now expand (20-2x)^2 and simplify to get a standard quadratic to solve.

\begin{aligned} (20-2x)^2&=(20-2x)(20-2x) \\ &=400-40x-40x+4x^2 \\ &=4x^2-80x+400 \end{aligned}

\begin{aligned} x^2+(4x^2-80x+400)&=80 \\5x^2-80x+400&=80 \end{aligned}

To solve this quadratic, we need one side (here the RHS) to equal zero, so subtract 80 from both sides,

5x^{2}-80x+320=0.

Notice that each coefficient is a multiple of 5, so we can divide through by 5 to make the numbers smaller and easier to deal with,

x^{2}-16x+64=0.

Finally, solve the quadratic. We do this by factorisation here, although you could use either of the other two methods (completing the square or the quadratic formula).

\begin{aligned} x^{2}-16x+64&=0 \\ (x-8)(x-8)&=0 \end{aligned}

We have one solution, x=8.

In situations like this one where you are told that a line is a tangent and asked to prove it, you should expect to see a repeated root and one solution, so look out for this when factorising.

**Make a statement about the solution.**

Because we are writing a proof, we need to end our mathematical working with a statement.

Here we write,

the line and the circle intersect at one point (x=8), so the line must be a tangent to the circle.

If you wanted to, you could substitute the x value into either equation to find the corresponding y value, and hence the coordinates of the point of intersection – but this isn’t necessary for this particular question.

**Using the gradient of the radius rather than the gradient of the tangent in the final equation**

Be careful that you don’t miss out a step – the gradient of the radius enables us to use perpendicularity to then find the gradient of the tangent.

**Dividing the wrong way round when calculating gradients**

Remember that the gradient is calculated as \cfrac{\text{ change in y}}{\text{change in x}}.

**Forgetting to change the sign when finding the perpendicular gradient**

The gradient of the tangent is the negative reciprocal of the gradient of the radius.

1. Find the equation of the tangent to the circle x^{2}+y^{2}=45 at the point (-6,3).

y=-\cfrac{1}{2} x-\cfrac{9}{2}

y=\cfrac{1}{2} x+\cfrac{15}{2}

y=2 x+15

y=2x-12

The gradient of the radius is -\cfrac{1}{2}, so the gradient of the tangent is 2.

Substitute x=-6 and y=3 into y=2x+c to get the answer y=2x+15.

2. Find the equation of the tangent to the circle x^{2}+y^{2}=40 at the point (-6,2).

y=3x-12

y=3x+20

y=-3x-16

y=\cfrac{1}{3} x+4

The gradient of the radius is -\cfrac{1}{3}, so the gradient of the tangent is 3.

Substitute x=-6 and y=2 into y=3x+c to get the answer y=3x+20.

3. Find the equation of the tangent to the circle x^{2}+y^{2}=52 at the point (6,-4).

y=-\cfrac{3}{2}x+5

y=\cfrac{3}{2}x-13

y=-\cfrac{3}{2}x+18

y=-\cfrac{2}{3}x-8

The gradient of the radius is -\cfrac{2}{3}, so the gradient of the tangent is \cfrac{3}{2}.

Substitute x=6 and y=-4 to get the answer y=\cfrac{3}{2}x-13.

4. Find the equation of the tangent to the circle x^{2}+y^{2}=50 at the point (1,-7).

Write your answer in the form ax+by+d=0.

7y-x+50=0

7y+x+48=0

y=\cfrac{1}{7} x+50

y+7x=0

The gradient of the radius is -7, so the gradient of the tangent is \cfrac{1}{7}.

Substitute x=1 and y=-7 to get the answer y=\cfrac{1}{7} x-\cfrac{50}{7},

then multiply by 7, giving 7y=x-50 and rearrange to the correct form.

5. The diagram shows the circle x^{2}+y^{2}=40.

The tangent to the circle at the point (2,6) intersects the x axis at point A.

Find the x coordinate of point A.

x=-16

x=20

x=12

x=40

The gradient of the radius is 3, so the gradient of the tangent is -\cfrac{1}{3}.

Substitute x=2 and y=6 to get the answer x+3y=20.

Then substitute y=0 to get x=20.

6. The diagram shows the circle x^{2}+y^{2}=100.

The tangent to the circle at the point (-8,6) intersects the x axis at point A.

Find the x coordinate of point A.

x=\cfrac{-32}{3}

x=-10

x=12

x=-\cfrac{25}{2}

The gradient of the radius is -\cfrac{3}{4}, so the gradient of the tangent is \cfrac{4}{3}.

Substitute x=-8 and y=6 to get the answer 4x-3y+50=0.

Then substitute y=0 to get x=-\cfrac{25}{2} or -12.5.

1. Here is a circle, centre O. The tangent to the circle at the point P(2,-4) is shown.

Find an equation of the tangent at the point P.

**(3 marks)**

Show answer

Gradient of radius =\cfrac{-4}{2}=-2

**(1)**

Gradient of tangent =\cfrac{1}{2}

**(1)**

Equation of tangent is y=\cfrac{1}{2} x-5 \ or \ 2x-4y=20

**(1)**

2. The line l is a tangent to the circle x^{2}+y^{2}=18 at the point A.

A is the point (3,3).

The line l crosses the x -axis at point B.

Work out the area of triangle OAB.

**(5 marks)**

Show answer

Gradient of radius =\cfrac{3}{3}=1

**(1)**

Gradient of tangent =-1

**(1)**

Equation of tangent is x+y=6, or finding c=6

**(1)**

x -coordinate of B is 6

**(1)**

Area of triangle =\cfrac{6 \times 3}{2}=9 (square units)

**(1)**

3. Prove algebraically that the straight line with equation 3y-x=30 is a tangent to the circle with equation x^{2}+y^{2}=90.

**(5 marks)**

Show answer

Rearrange 3y-x=30 \ to \ x=3y-30 and substitute into x^{2}+y^{2}=90.

**(1)**

**(1)**

Simplify the equation of the circle to

10 y^{2}-180 y+810=0 \Rightarrow y^{2}-18 y+81=0.**(1)**

Solve to get y=9.

**(1)**

Statement: There is only one point of intersection so the line is a tangent to the circle.

**(1)**

You have now learned how to:

- Find the equation of the tangent to a circle at a given point
- Find the x -coordinate of the point where the tangent intersects the x -axis
- Prove algebraically that a given line is a tangent to a given circle

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