Finding the equation of a radius of a circle
I can find the equation of a radius of a circle.
Finding the equation of a radius of a circle
I can find the equation of a radius of a circle.
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.
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).
Video
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Starter quiz
6 Questions
Exit quiz
6 Questions
$$(3,5)$$ -
$$(x-3)^2+(y-5)^2=1$$
$$(-3,5)$$ -
$$(x+3)^2+(y-5)^2=1$$
$$(-3,-5)$$ -
$$(x+3)^2+(y+5)^2=1$$
$$(5,-3)$$ -
$$(x-5)^2+(y+3)^2=1$$
$$(5,3)$$ -
$$(x-5)^2+(y-3)^2=1$$
$$(-5,3)$$ -
$$(x+5)^2+(y-3)^2=1$$