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
Starter quiz
6 Questions
$$(7,8)$$ -
$$(x-7)^2+(y-8)^2=25$$
$$(7,-8)$$ -
$$(x-7)^2+(y+8)^2=25$$
$$(-7,8)$$ -
$$(x+7)^2+(y-8)^2=25$$
$$(8,7)$$ -
$$(x-8)^2+(y-7)^2=25$$
$$(-8,7)$$ -
$$(x+8)^2+(y-7)^2=25$$
$$(8,-7)$$ -
$$(x-8)^2+(y+7)^2=25$$