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Year 11
Higher

Finding the equation of the tangent to a circle

I can find the equation of the tangent to a circle at any given point.

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New
New
Year 11
Higher

Finding the equation of the tangent to a circle

I can find the equation of the tangent to a circle at any given point.

Download all resources
Share activities with pupils

Link copied to clipboard

These resources will be removed by end of Summer Term 2025.

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Lesson details

Key learning points

  1. Using the gradient of the radius through a given point, you can find the equation of the tangent at this same point.
  2. You have already proved that the tangent at any point on a circle is perpendicular to the radius at that point.
  3. You have already proved that the product of the gradients of two perpendicular lines is -1
  4. Using the gradient of the tangent and the coordinates of the point, you can find the equation of the tangent.

Keywords

  • Gradient - The gradient is a measure of how steep a line is. It is calculated by finding the rate of change in the y-direction with respect to the positive x-direction.

  • Radius - The radius is any line segment that joins the centre of a circle to its edge.

  • Tangent - A tangent of a circle is a line that intersects the circle exactly once.

Common misconception

Pupils can confuse the gradient of the radius with the length of the radius.

A sketch will help pupils apply the right skills. To find the equation of a straight line we need the gradient. You may wish to draw concentric circles to show pupils that the radii are different lengths but can have same gradient.


To help you plan your year 11 maths lesson on: Finding the equation of the tangent to a circle, download all teaching resources for free and adapt to suit your pupils' needs...

Pupils should be getting used to fractional answers. They should be checking their answers by substituting known coordinates into their equations. Graphing software will also help pupils check answers.
Teacher tip

Licence

This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Lesson video

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Starter quiz

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6 Questions

Q1.
The tangent at any point on a circle __________ to the radius at that point.
Correct answer: is at a right angle
has an acute angle
is parallel
Correct answer: is perpendicular
is at an alternate angle
Q2.
What is the gradient of the line passing through the points $$(-7,-8)$$ and $$(-9,10)$$?
$$9$$
$$1\over9$$
$$-{1\over9}$$
Correct answer: $$-9$$
Q3.
What is the equation of the line passing through the points $$(8,11)$$ and $$(-12,6)$$?
$$y=4x-9$$
$$y=4x+9$$
$$y={1\over4}x-9$$
$$y=9-{1\over4}x$$
Correct answer: $$y={1\over4}x+9$$
Q4.
Match the centre of each circle to the equation for the circle.
Correct Answer:$$(7,8)$$,$$(x-7)^2+(y-8)^2=25$$

$$(x-7)^2+(y-8)^2=25$$

Correct Answer:$$(7,-8)$$,$$(x-7)^2+(y+8)^2=25$$

$$(x-7)^2+(y+8)^2=25$$

Correct Answer:$$(-7,8)$$,$$(x+7)^2+(y-8)^2=25$$

$$(x+7)^2+(y-8)^2=25$$

Correct Answer:$$(8,7)$$,$$(x-8)^2+(y-7)^2=25$$

$$(x-8)^2+(y-7)^2=25$$

Correct Answer:$$(-8,7)$$,$$(x+8)^2+(y-7)^2=25$$

$$(x+8)^2+(y-7)^2=25$$

Correct Answer:$$(8,-7)$$,$$(x-8)^2+(y+7)^2=25$$

$$(x-8)^2+(y+7)^2=25$$

Q5.
What is the gradient of the radius connecting the centre of the circle $$(x-175)^2+(y+65)^2=250$$ to the point $$(190,-70)$$ on the circumference?
$$3$$
$$1\over3$$
Correct answer: $$-{1\over3}$$
$$-3$$
Q6.
What is the equation of the radius connecting the centre of the circle $$(x-5)^2+(y+8)^2=10$$ to the point $$(4,-5)$$ on the circumference?
$$y=3-7x$$
Correct answer: $$y=7-3x$$
$$y=7+3x$$
$$y=3x-7$$
$$y=-3x-7$$

Exit quiz

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6 Questions

Q1.
A tangent of a circle is a line that intersects the circle __________.
at its centre
twice
Correct answer: exactly once
Q2.
From the equation: $$(x-7)^2+(y-10)^2=25$$ we can deduce what?
Correct answer: The $$x$$-coordinate of the centre is $$7$$
The $$x$$-coordinate of the centre is $$-7$$
Correct answer: The $$y$$-coordinate of the centre is $$10$$
The $$y$$-coordinate of the centre is $$-10$$
The radius is $$25$$
Q3.
The equation of a circle is $$(x-2)^2+(y-3)^2=80$$ Which of the below coordinate pairs will be on the radius connecting the centre of the circle to the point $$(6,11)$$ on the circumference?
$$(5,3)$$
Correct answer: $$(3,5)$$
$$(1,1)$$
Correct answer: $$(5,9)$$
$$(8,15)$$
Q4.
The gradient of the radius connecting a circle's centre to the point $$(4,11)$$ is $$4$$. What is the gradient of the tangent to the circle at this point?
$$4$$
$$-4$$
$${1\over4}$$
Correct answer: $$-{1\over4}$$
$$-{4\over11}$$
Q5.
The gradient of the radius connecting a circle's centre to the point $$(6,6)$$ is $$-{1\over3}$$. What is the equation of the tangent to the circle at this point? Answer in the form $$y=mx+c$$
Correct Answer: y=3x-12, y=3x+-12, y=3x+(-12), y = 3x + -12, y = 3x + (-12)
Q6.
What is the equation of the tangent to the circle $$(x+15)^2+(y-8)^2=13$$ at the point $$(-13,5)$$?
$$y={-{3\over2}}x+{-{29\over2}}$$
$$y={3\over2}x+{41\over3}$$
Correct answer: $$y={2\over3}x+{41\over3}$$
$$y={-{3\over2}}x+{41\over3}$$
$$y={2\over3}x-{41\over3}$$