G8 Develop the equation of a circle centred at the origin

Ontario Grade 10 Mathematics

Analytic Geometry

Use the coordinates of a point and its distance from the centre to build the equation.

A circle is the set of points that are all the same distance from one fixed point. That fixed point is the centre. The distance from the centre to the circle is the radius. In this lesson, the centre is the origin,

(0,0)

. We will use familiar ideas about coordinates, squares, and right triangles to develop an equation for every point on the circle.

What you will learn

Describe a circle centred at the origin using its radius.
Connect a point on a circle to a right triangle on a coordinate grid.
Develop the equation $x^2+y^2=r^2$ for a circle centred at the origin.
Write and check the equation when the radius is known.

1. Grade 9 bridge: coordinates and distance

A coordinate grid has a horizontal

x

-axis and a vertical

y

-axis. They meet at the origin, whose coordinates are

(0,0)

. A point such as

(3,4)

is 3 units to the right of the origin and 4 units up.

The horizontal and vertical moves form the two shorter sides of a right triangle. The straight-line distance between the origin and the point is the longest side. The Pythagorean theorem connects the lengths of these sides: the squares of the shorter sides add to the square of the longest side.

For the point

(3,4)

, the shorter sides have lengths 3 and 4. Their squared lengths add to 25, so the straight-line distance is 5. This familiar relationship is the key to describing a circle.

a^2+b^2=c^2

The origin is $(0,0)$ .
The coordinates give horizontal and vertical distances from the axes.
For a right triangle, the squared lengths of the shorter sides add to the square of the longest side.

2. From a circle to a rule

Imagine a circle centred at the origin with radius

r

. The letter

r

stands for the fixed distance from the centre to any point on the circle. Choose any point on the circle and call its coordinates

(x,y)

. The letters

x

and

y

represent the point’s horizontal and vertical coordinates.

Draw a line from the origin to

(x,y)

. Its length is

r

, because the point lies on the circle. The horizontal and vertical sides of the right triangle have lengths based on

x

and

y

. A coordinate can be negative, but squaring it gives the same result as squaring its distance from zero. So the squared side lengths are

x^2

and

y^2

.

Apply the Pythagorean theorem to this triangle. The sum of the squared horizontal and vertical distances equals the squared radius. This is not a rule about points inside or outside the circle: it describes points exactly on the circle. A point inside is closer to the centre; a point outside is farther away.

The equation works for every point on the circle. For example, when a point has a negative coordinate, its square is still positive. This lets the same equation describe points in all four parts of the grid.

x^2+y^2=r^2

The point $(x,y)$ is on the circle, so its distance from $(0,0)$ is $r$ .
The horizontal and vertical distances contribute $x^2$ and $y^2$ .
The circle equation is the condition that a point’s distance from the origin is exactly the radius.

3. Read the equation and build one

The equation has a simple meaning: square the point’s

x

-coordinate, square its

y

-coordinate, and add the results. For a point on the circle, the total must equal the square of the radius. If the radius is known, substitute its value for

r

and calculate

r^2

.

For instance, a circle with radius 5 has a right-hand side of 25. Its equation is

x^2+y^2=25

. The equation names a whole circle, not one point. Many coordinate pairs can make the left side equal 25.

To check whether a given point is on a circle, substitute its coordinates for

x

and

y

. If the left side equals the right side, the point is on the circle. If it does not, the point is not on the circle. Keep the coordinates in their correct places, then square and add.

You can also develop the equation when you know a point on the circle instead of its radius. Use the point’s coordinates in the left side, find the total, and set that total equal to

r^2

. If that total is a perfect square, its square root gives the radius.

r^2=x^2+y^2

Substitute the radius, then square it to find the constant on the right.
Substitute a point’s coordinates to check whether it lies on the circle.
Do not add the coordinates before squaring them.

4. Independent practice and a final check

Try these without looking at the worked example first. For a circle centred at the origin with radius 6, write the equation. Then decide whether

(0,-6)

lies on it. For a separate circle, a point on it is

(-5,12)

. Use the point to find

r^2

and the radius.

For the first task, remember that the constant on the right is the square of the radius, not the radius itself. For the second task, substitute both coordinates into the left side. Negative coordinates still get squared. Compare your result with the right side, and use the positive distance as the radius.

When you finish, check that your equation has both squared coordinate terms and that its right side is the squared radius. Those features show that you have used the distance relationship for a circle centred at the origin.

Use the radius to develop the equation.
Use a point on the circle to find the squared radius.
Check by comparing both sides of the equation.

How a point on the circle fits the equation

Point on circle	Squared horizontal coordinate	Squared vertical coordinate	Sum
$(3,4)$	$3^2=9$	$4^2=16$	$9+16=25$
$(-3,4)$	$(-3)^2=9$	$4^2=16$	$9+16=25$
$(0,5)$	$0^2=0$	$5^2=25$	$0+25=25$

Worked example

Develop an equation from a point on the circle

A circle is centred at the origin. The point

(3,4)

lies on the circle. Develop the equation of the circle.

Name the coordinates
The point on the circle has horizontal coordinate 3 and vertical coordinate 4. Its distance from the origin is the radius, so we can use these side lengths to find the squared radius.
$(x,y)=(3,4)$
Find the squared radius
The Pythagorean theorem says to square each shorter side and add. The result is the square of the distance from the origin, which is $r^2$ .
$r^2=3^2+4^2=9+16=25$
Write the circle equation
Use the squared radius as the right side of the circle equation. This equation includes every point whose squared horizontal and vertical distances add to 25.
$x^2+y^2=25$

Answer: The equation is

x^2+y^2=25

. The radius is 5 units.

Check: Substitute the known point:

3^2+4^2=9+16=25

. It satisfies the equation, as it should.

Common mistakes and how to avoid them

Using the radius itself on the right, such as writing

x^2+y^2=5

for a radius of 5.

Correction: Square the radius. A radius of 5 gives

r^2=25

, so the equation is

x^2+y^2=25

.

Adding the coordinates first, as in

(x+y)^2=r^2

.

Correction: The two sides of the right triangle are the horizontal and vertical distances. Square each coordinate separately, then add:

x^2+y^2=r^2

.

Treating a negative coordinate as a negative distance after squaring.

Correction: A negative coordinate becomes positive when squared. For example,

(-3)^2=9

.

Assuming a point that does not satisfy the equation is still on the circle.

Correction: A point is on the circle only when its squared coordinates add to the squared radius.

Lesson summary

A circle centred at the origin has centre $(0,0)$ and a fixed radius $r$ .
The Pythagorean theorem connects a point’s coordinates to its distance from the origin.
The equation of the circle is $x^2+y^2=r^2$ .
Substitute a known radius or a point on the circle to develop the equation.

Check your understanding

Question 1

What is the equation of a circle centred at the origin with radius 4?

$x^2+y^2=4$
$x^2+y^2=8$
$x^2+y^2=16$
$x+y=16$

Show answer and explanation

x^2+y^2=16

The right side is the radius squared. Since

4^2=16

, the equation is

x^2+y^2=16

.

Question 2

Does the point

(-6,8)

lie on the circle

x^2+y^2=100

?

Yes, because $(-6)^2+8^2=100$ .
No, because the first coordinate is negative.
No, because $-6+8=2$ .
Yes, because $(-6)^2+8^2=28$ .

Show answer and explanation

Yes, because

(-6)^2+8^2=100

.

Squaring gives

36+64=100

, which matches the right side. The point is on the circle.

Question 3

A point

(5,12)

lies on a circle centred at the origin. What is the radius?

13 units
17 units
119 units
169 units

Show answer and explanation

13 units

The squared radius is

5^2+12^2=25+144=169

. The radius is the positive distance whose square is 169, so it is 13 units.

Key terms

Origin: The point $(0,0)$ where the horizontal and vertical axes meet.
Circle: The set of points that are all the same distance from one fixed centre.
Radius: The distance from the centre of a circle to any point on the circle.
Equation: A mathematical statement showing that two expressions have the same value.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic G8. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

G8 · Develop the equation of a circle centred at the origin

Ontario Grade 10 Mathematics

Use the coordinates of a point and its distance from the centre to build the equation.

What you will learn

1. Grade 9 bridge: coordinates and distance

2. From a circle to a rule

3. Read the equation and build one

4. Independent practice and a final check

How a point on the circle fits the equation

Develop an equation from a point on the circle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MPM2D

About this lesson