Finding the equation of the tangent to a circle | Maths

Starter quiz

The tangent at any point on a circle __________ to the radius at that point.

is at a right angle ✓

has an acute angle

is parallel

is perpendicular ✓

is at an alternate angle
What is the gradient of the line passing through the points $(-7,-8)$ and $(-9,10)$ ?

$9$

$1\over9$

$-{1\over9}$

$-9$ ✓
What is the equation of the line passing through the points $(8,11)$ and $(-12,6)$ ?

$y=4x-9$

$y=4x+9$

$y={1\over4}x-9$

$y=9-{1\over4}x$

$y={1\over4}x+9$ ✓
Match the centre of each circle to the equation for the circle.

$(7,8)$

⇔

$(x-7)^2+(y-8)^2=25$ ✓

$(7,-8)$

⇔

$(x-7)^2+(y+8)^2=25$ ✓

$(-7,8)$

⇔

$(x+7)^2+(y-8)^2=25$ ✓

$(8,7)$

⇔

$(x-8)^2+(y-7)^2=25$ ✓

$(-8,7)$

⇔

$(x+8)^2+(y-7)^2=25$ ✓

$(8,-7)$

⇔

$(x-8)^2+(y+7)^2=25$ ✓
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?

$3$

$1\over3$

$-{1\over3}$ ✓

$-3$
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?

$y=3-7x$

$y=7-3x$ ✓

$y=7+3x$

$y=3x-7$

$y=-3x-7$

Exit quiz

A tangent of a circle is a line that intersects the circle __________.

at its centre

twice

exactly once ✓
From the equation: $(x-7)^2+(y-10)^2=25$ we can deduce what?

The $x$ -coordinate of the centre is $7$ ✓

The $x$ -coordinate of the centre is $-7$

The $y$ -coordinate of the centre is $10$ ✓

The $y$ -coordinate of the centre is $-10$

The radius is $25$
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?

$(5,3)$

$(3,5)$ ✓

$(1,1)$

$(5,9)$ ✓

$(8,15)$
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?

$4$

$-4$

${1\over4}$

$-{1\over4}$ ✓

$-{4\over11}$
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 $y=mx+c$

'y=3x-12' ✓
What is the equation of the tangent to the circle $(x+15)^2+(y-8)^2=13$ at the point $(-13,5)$ ?

$y={-{3\over2}}x+{-{29\over2}}$

$y={3\over2}x+{41\over3}$

$y={2\over3}x+{41\over3}$ ✓

$y={-{3\over2}}x+{41\over3}$

$y={2\over3}x-{41\over3}$

Worksheet

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Presentation

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Video

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.

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.

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.