Finding the equation of the tangent to a circle
I can find the equation of the tangent to a circle at any given point.
Finding the equation of the tangent to a circle
I can find the equation of the tangent to a circle at any given point.
These resources will be removed by end of Summer Term 2025.
Lesson details
Key learning points
- Using the gradient of the radius through a given point, you can find the equation of the tangent at this same point.
- You have already proved that the tangent at any point on a circle is perpendicular to the radius at that point.
- You have already proved that the product of the gradients of two perpendicular lines is -1
- 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.
Licence
This content is © Oak National Academy Limited (2024), 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
6 Questions
$$(7,8)$$ -
$$(x-7)^2+(y-8)^2=25$$
$$(7,-8)$$ -
$$(x-7)^2+(y+8)^2=25$$
$$(-7,8)$$ -
$$(x+7)^2+(y-8)^2=25$$
$$(8,7)$$ -
$$(x-8)^2+(y-7)^2=25$$
$$(-8,7)$$ -
$$(x+8)^2+(y-7)^2=25$$
$$(8,-7)$$ -
$$(x-8)^2+(y+7)^2=25$$