Equation Of Tangent

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.

What is the equation of a tangent?

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}

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.

How to find the equation of a tangent

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

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

Explain how to find the equation of a tangent

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:

Equation of tangent examples

Example 1: finding the equation of a tangent

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

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

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

2Find the gradient of the tangent at that point.

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

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

3Find 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.

Example 2: finding the equation of a tangent

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

Find the gradient of the tangent at that point.

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

Example 3: dealing with trickier fractions

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

Find the gradient of the tangent at that point.

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

Example 4: finding the point of intersection on the x axis

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 tangent at that point.

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

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

Example 5: finding the equation of a tangent with a non-zero centre

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 tangent at that point.

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

Proof using equation of a tangent

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.

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.

How to prove that a line is a tangent to a circle

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

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

Proof using equation of tangent examples

Example 6: proving that a line is a tangent to a circle

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.

Simplify and solve the quadratic equation.

Make a statement about the solution.

Common misconceptions

• 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

Practice equation of tangent questions

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

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

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

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).

7y-x+50=0

7y+x+48=0

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

y+7x=0

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

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}

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.

Equation of tangent GCSE questions

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)

(1)

(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)

(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)

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

(1)

(3y-30)^{2}=9 y^{2}-180 y+900

(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)

Learning checklist

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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