Equation of a Circle: Centre at Origin Revision Notes for Grade 12 NSC Matric Mathematics

Equation of a Circle: Centre at Origin

What is a circle?

A circle is the set of all points that are an equal distance (radius) from a given point (centre). In other words, every point on the circumference of a circle is equidistant from its centre.

The radius of a circle is the distance from the centre of a circle to any point on the circumference. This distance remains constant for all points on the circle.

A diameter of a circle is any line passing through the centre of the circle which connects two points on the circle. The diameter is also the maximum distance between two points on a circle.

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Understanding these fundamental definitions is crucial for working with circles. Remember that the radius is the key measurement that determines the size of a circle.

Deriving the equation of a circle with centre at the origin

When a circle has its centre at the origin O(0; 0), we can derive its equation using the Pythagorean theorem.

Consider any point P(x; y) on the circumference of a circle with radius r and centre at O(0; 0).

In triangle OPQ (where Q is the foot of the perpendicular from P to the x-axis):

OP = r (radius of the circle)
PQ = y - 0 = y (vertical distance)
OQ = x - 0 = x (horizontal distance)

Using the Pythagorean theorem: $OP^2 = PQ^2 + OQ^2$

Substituting the values: $r^2 = y^2 + x^2$

Rearranging: $x^2 + y^2 = r^2$

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This is the standard form equation of a circle with centre at the origin.

Key formula

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Equation of a circle with centre at the origin:

If P(x; y) is a point on a circle with centre O(0; 0) and radius r, then the equation of the circle is:

$x^2 + y^2 = r^2$

Circle symmetry

A circle with centre at the origin has special symmetry properties:

Origin symmetry: For every point (x; y) on the circumference, the point (-x; -y) is also on the circle
Axis symmetry: A circle centred on the origin is symmetrical about both the x-axis and y-axis
Line symmetry: The circle is also symmetrical about the lines y = x and y = -x

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This means a circle has infinitely many lines of symmetry - any line passing through the centre is a line of symmetry.

Worked example 1: Basic circle equation

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Worked Example: Basic Circle Equation

Question: Given a circle with centre O(0; 0) and radius of 3 units.

Sketch the circle on the Cartesian plane
Determine the equation of the circle
Show that the point T(-√4; √5) lies on the circle

Solution:

Step 1: Draw a sketch

Step 2: Determine the equation

Using the general form: $x^2 + y^2 = r^2$

Substitute r = 3: $x^2 + y^2 = (3)^2$

Therefore: $x^2 + y^2 = 9$

Step 3: Show that point T lies on the circle

Substitute the coordinates of T(-√4; √5) into the equation:

LHS = $x^2 + y^2$

= $(-\sqrt{4})^2 + (\sqrt{5})^2$

= $4 + 5$

= $9$

= $r^2$

= RHS

Therefore, T(-√4; √5) lies on the circle $x^2 + y^2 = 9$ .

Worked example 2: Circle through two points

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Worked Example: Circle Through Two Points

Question: A circle with centre O(0; 0) passes through the points P(-5; 5) and Q(5; -5).

Plot the points and draw a rough sketch
Determine the equation of the circle
Calculate the length of PQ
Explain why PQ is a diameter

Solution:

Step 1: Draw a sketch

Step 2: Determine the equation

Using the general form $x^2 + y^2 = r^2$ and substituting P(-5; 5):

$(-5)^2 + (5)^2 = r^2$

$25 + 25 = r^2$

$50 = r^2$

Therefore $r = \sqrt{50} = 5\sqrt{2}$ units

The equation is: $x^2 + y^2 = 50$

Step 3: Calculate the length PQ

Using the distance formula:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$= \sqrt{(5 - (-5))^2 + (-5 - 5)^2}$

$= \sqrt{(10)^2 + (-10)^2}$

$= \sqrt{100 + 100}$

$= \sqrt{200}$

$= 10\sqrt{2}$ units

Step 4: Why PQ is a diameter

The radius $r = 5\sqrt{2}$

The diameter $d = 2 \times r = 2 \times 5\sqrt{2} = 10\sqrt{2}$

Since $PQ = 10\sqrt{2} = d$ , PQ is a diameter of the circle.

Alternative method: Using symmetry, P(-5; 5) lies opposite Q(5; -5) with respect to the origin, so PQ passes through the centre and is therefore a diameter.

Worked example 3: Finding points with specific conditions

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Worked Example: Finding Points with Specific Conditions

Question: Given a circle with centre O(0; 0) and radius √45 units. Determine the possible coordinates of the points on the circle which have an x-value that is twice the y-value.

Solution:

Step 1: Determine the equation

$x^2 + y^2 = r^2$

$x^2 + y^2 = (\sqrt{45})^2$

$x^2 + y^2 = 45$

Step 2: Find coordinates where x = 2y

Substitute $x = 2y$ into the equation:

$(2y)^2 + y^2 = 45$

$4y^2 + y^2 = 45$

$5y^2 = 45$

$y^2 = 9$

$y = \pm 3$

When $y = -3$ : $x = 2(-3) = -6$

When $y = 3$ : $x = 2(3) = 6$

Therefore, the points are (6; 3) and (-6; -3).

Check:

For (6; 3): $6^2 + 3^2 = 36 + 9 = 45$ ✓

For (-6; -3): $(-6)^2 + (-3)^2 = 36 + 9 = 45$ ✓

Exam tips

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Key Exam Strategies:

Always start with a sketch when solving circle problems - it helps visualise the situation
Remember the standard form: $x^2 + y^2 = r^2$ for circles with centre at origin
Use symmetry properties to check your answers or find alternative solutions
Substitute coordinates into the equation to verify if points lie on the circle
The radius is always positive - take the positive square root when finding r
Diameter = 2 × radius - use this relationship in problems involving diameters

Remember!

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Key Points to Remember:

A circle is the set of all points equidistant from the centre
The equation of a circle with centre at origin is $x^2 + y^2 = r^2$
Use the Pythagorean theorem to derive the circle equation
Circles with centre at origin have multiple lines of symmetry
Always substitute coordinates to verify if points lie on the circle

Equation of a Circle: Centre at Origin (Grade 12 NSC Matric Mathematics): Revision Notes