Finding the equation of a radius of a circle | Maths | Quizalize & Oak National Academy

Starter quiz

The radius is any __________ that joins the centre of a circle to its edge.

diameter

line segment ✓

arc

chord

circumference
What is the gradient of this line?

'6' ✓
What is the gradient of the line passing through coordinate pairs $(-11,11)$ and $(21,-53)$ ?

'-2' ✓
What is the gradient of the line passing through coordinate pairs $(-11,11)$ and $(21,-125)$ ?

${{32}\over{-136}}$

${{-17}\over{4}}$ ✓

${{-32}\over{136}}$

${{17}\over{4}}$

${{-4}\over{17}}$
What is the equation of the line passing through coordinate pairs $(-10,-57)$ and $(15,68)$ ?

$y=5x-7$ ✓

$y=5x+7$

$y=7 - 5x$

$y=-5x-107$

$y=5x+107$
Which of the below equations will form a circle?

$x^2+y=9^2$

$x^2+y^2=9$ ✓

$x^2-y^2=9$

$x^2+y^2=9^2$ ✓

$(x+y)^2=9$

Exit quiz

In the equation of this circle $(x-a)^2+(y-b)^2=r^2$ the value of $r$ affects the __________ of the circle.

radius ✓

ratio

range

rate of change

centre
In the equation of this circle $(x-a)^2+(y-b)^2=r^2$ the values of $a$ and $b$ affect the __________ of the circle.

chord

circumference

radius

diameter

centre ✓
Match the coordinate pairs for the centre of each circle to the correct equation for that circle.

$(3,5)$

⇔

$(x-3)^2+(y-5)^2=1$ ✓

$(-3,5)$

⇔

$(x+3)^2+(y-5)^2=1$ ✓

$(-3,-5)$

⇔

$(x+3)^2+(y+5)^2=1$ ✓

$(5,-3)$

⇔

$(x-5)^2+(y+3)^2=1$ ✓

$(5,3)$

⇔

$(x-5)^2+(y-3)^2=1$ ✓

$(-5,3)$

⇔

$(x+5)^2+(y-3)^2=1$ ✓
What is the gradient of the radius connecting the centre of the circle $(x-8)^2+(y-2)^2=10$ to the point $(9,5)$ on the circumference?

$3$ ✓

$1\over3$

$-1\over3$

$-3$
What is the equation of the radius connecting the centre of the circle $(x+5)^2+(y-3)^2=5$ to the point $(-4,5)$ on its circumference? Write your answer in the form $y=mx+c$

'y=2x+13' ✓
What is the equation of the radius connecting the centre of the circle $(x+5)^2+(y+8)^2=20$ to the point $(-1,-6)$ on its circumference?

$y={1\over2}x-{11\over2}$ ✓

$y=-{1\over2}x-{11\over2}$

$y={1\over2}x+{11\over2}$

$y={1\over2}x-{2\over11}$

$y=2x-{11\over2}$

Worksheet

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Lesson Details

Key learning points

The equation of a circle gives the coordinates of the centre of the circle.
Using this and the coordinates of a point on the circle, you can calculate the gradient of the radius.
Using the gradient and the centre of the circle, you can find the equation of this radius.

Common misconception

Pupils may think that a circle with equation $(x+a)^2 + (y+b)^2=r^2$ has centre $(a,b)$

The general form of an equation of a circle is $(x-a)^2 + (y-b)^2=r^2$ where $(a,b)$ is then the centre. Graphing software will be useful to show and explore this general form.

Keywords

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