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

Equation of a Circle: Centre at (a;b)

What is the equation of a circle with centre at (a;b)?

When a circle is not centred at the origin, we need to adjust our equation to account for the shifted centre position. Understanding this concept is fundamental to solving many coordinate geometry problems involving circles.

infoNote

Definition: A circle is the set of all points that are equidistant from a fixed point called the centre. The distance from the centre to any point on the circle is called the radius.

Deriving the general equation

To find the equation of a circle with centre at point C(a; b), we use the distance formula. Consider any point P(x; y) on the circumference of the circle.

The distance from the centre C(a; b) to any point P(x; y) on the circle equals the radius r:

Distance formula: $PC = \sqrt{(x - a)^2 + (y - b)^2}$

Since this distance equals the radius: $r = \sqrt{(x - a)^2 + (y - b)^2}$

Squaring both sides to eliminate the square root: $r^2 = (x - a)^2 + (y - b)^2$

Therefore, the standard form of a circle equation is:

chatImportant

Key Formula: $(x - a)^2 + (y - b)^2 = r^2$

Where:

(a; b) is the centre of the circle
r is the radius
(x; y) represents any point on the circle

Special case: Circle centred at the origin

When the centre is at the origin (0; 0), the equation becomes: $(x - 0)^2 + (y - 0)^2 = r^2$

This simplifies to: $x^2 + y^2 = r^2$

Worked example 1: Finding the equation from centre and point

lightbulbExample

Worked Example: Finding the equation from centre and point

Question: F(6; -4) is a point on the circle with centre C(3; -4). Find the equation of the circle.

Solution:

Step 1: Write the general equation

$(x - a)^2 + (y - b)^2 = r^2$

Step 2: Substitute the centre coordinates (3; -4)

$(x - 3)^2 + (y - (-4))^2 = r^2$

$(x - 3)^2 + (y + 4)^2 = r^2$

Step 3: Find the radius using point F(6; -4)

Substitute F(6; -4) into the equation:

$(6 - 3)^2 + (-4 + 4)^2 = r^2$

$(3)^2 + (0)^2 = r^2$

$9 = r^2$

Step 4: Write the final equation

$(x - 3)^2 + (y + 4)^2 = 9$

Worked example 2: Converting from general form to standard form

lightbulbExample

Worked Example: Converting from general form to standard form

Question: Find the centre and radius of the circle: $3x^2 + 6x + 3y^2 - 12y - 33 = 0$

Solution:

Step 1: Make coefficients of $x^2$ and $y^2$ equal to 1

Divide everything by 3: $x^2 + 2x + y^2 - 4y - 11 = 0$

Step 2: Complete the square for x terms

Take half of the x coefficient: $(2/2)^2 = 1^2 = 1$

$x^2 + 2x + 1 - 1 + y^2 - 4y - 11 = 0$

$(x + 1)^2 - 1 + y^2 - 4y - 11 = 0$

Step 3: Complete the square for y terms

Take half of the y coefficient: $(-4/2)^2 = (-2)^2 = 4$

$(x + 1)^2 - 1 + (y - 2)^2 - 4 - 11 = 0$

$(x + 1)^2 + (y - 2)^2 = 16$

Answer: Centre is (-1; 2) and radius is 4 units.

Worked example 3: Circle transformations

lightbulbExample

Worked Example: Circle transformations

Question: A circle has equation $x^2 + y^2 = 16$ . If it is shifted 2 units down and 1 unit right, write the equation of the shifted circle.

Solution:

Step 1: Identify the original circle

Centre: (0; 0), Radius: 4

Step 2: Apply the transformations

Vertical shift: 2 units down means replace y with (y + 2)
Horizontal shift: 1 unit right means replace x with (x - 1)

Step 3: Write the new equation

$(x - 1)^2 + (y + 2)^2 = 16$

The new centre is (1; -2) with radius 4 units.

Circle transformations summary

infoNote

Key transformation rules:

Horizontal shift h units right: replace x with (x - h)
Vertical shift k units up: replace y with (y - k)
Combined shifts: $(x - h)^2 + (y - k)^2 = r^2$ has centre (h; k)

Worked example 4: Advanced problem with geometric constraints

lightbulbExample

Worked Example: Advanced problem with geometric constraints

Question: Given points S(-3; 4) and T(-3; -4), find the equation of a circle where these points are endpoints of a diameter.

Solution:

Step 1: Find the centre (midpoint of diameter)

Centre = $\left(\frac{x\_1 + x\_2}{2}, \frac{y\_1 + y\_2}{2}\right)$

Centre = $\left(\frac{-3 + (-3)}{2}, \frac{4 + (-4)}{2}\right) = (0; 0)$

Step 2: Calculate the radius

Radius = distance from centre to either point

$r = \sqrt{(-3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 3: Write the equation

$x^2 + y^2 = 25$

Important property: Any angle subtended by a diameter at the circumference is 90°.

Circle relationships

Understanding how circles interact is important for solving complex problems:

infoNote

Key relationships (where d = distance between centres, $r\_1$ and $r\_2$ = radii):

Concentric circles: Same centre, different radii
Internal tangent: $d = |r\_2 - r\_1|$ (circles touch internally)
Intersecting: $d < r\_1 + r\_2$ (circles cross at two points)
External tangent: $d = r\_1 + r\_2$ (circles touch externally)
Separate: $d > r\_1 + r\_2$ (circles don't touch)

Important exam tips

Always sketch the circle first to visualise the problem
Remember that (a; b) in the equation means the centre is at (a; b)
When completing the square, add and subtract the same value
Check your answer by substituting a known point back into the equation
For transformations, be careful with positive and negative signs

chatImportant

Common exam mistakes to avoid

Confusing the signs: $(x - a)^2$ means centre has x-coordinate +a
Forgetting to square both sides when using the distance formula
Mixing up horizontal and vertical shifts in transformations
Not simplifying the radius when $r^2$ is given as a perfect square

bookmarkSummary

Key Points to Remember:

Circle equation with centre (a; b): $(x - a)^2 + (y - b)^2 = r^2$
Distance from centre to any point on circle equals radius: Use this to find $r^2$
Completing the square: Take half the coefficient, square it, then add and subtract
Transformations: Horizontal shift h units right → replace x with (x - h); Vertical shift k units up → replace y with (y - k)
Always sketch the circle to check your solution makes geometric sense

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