The Equation of a Circle (Centre (a, b)) Revision Notes | GCSE AQA Higher Further Maths

1. Why the Circle Equation Works

Every point on a circle is exactly the same distance from the centre. That single idea is the engine behind the whole equation.

You already know that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. A circle with centre $(a, b)$ and radius $r$ is the set of all points $(x, y)$ where that distance equals $r$.

Setting the distance formula equal to $r$ and squaring both sides gives the equation you need: $(x - a)^2 + (y - b)^2 = r^2$. No memorisation required — the formula follows directly from Pythagoras.

2. The Formula and What Each Part Means

The trickiest part of this formula is the sign convention. The centre coordinates appear with their signs reversed inside the brackets. Centre $(3, -5)$ gives $(x - 3)^2 + (y + 5)^2 = r^2$, not $(x + 3)^2 + (y - 5)^2$.

3. Writing the Equation Given Centre and Radius

When an exam question asks you to write down the equation of a circle, you substitute $a$, $b$, and $r$ directly into the formula. The key step is squaring $r$ before writing the right-hand side.

Worked example: Writing the Equation of a Circle

Write down the equation of the circle with centre $(−2, 5)$ and radius $7$.

1. $(x - a)^2 + (y - b)^2 = r^2$ — Start by writing the standard formula — this earns a method mark [M1].
2. $a = -2, \quad b = 5, \quad r = 7$ — Read these directly from the question. Keep the negative sign with $a$.
3. $(x - (-2))^2 + (y - 5)^2 = 7^2$ — Replace $a$, $b$, $r$ with their values. Show the substitution clearly [M1].
4. $(x + 2)^2 + (y - 5)^2 = 49$ — $x - (-2) = x + 2$ and $7^2 = 7 \times 7 = 49$. This is the final equation [A1].

Answer: $(x + 2)^2 + (y - 5)^2 = 49$

Notice how a negative $x$-coordinate in the centre turned into a plus sign in the bracket. This is the most common place to drop a mark, so always double-check the sign after substituting.

4. Reading the Centre and Radius from an Equation

The reverse skill — extracting the centre and radius from a given equation — is equally common in exams. The process is a two-step read: identify $a$ and $b$ from the brackets, then square-root the right-hand side to find $r$.

Worked example: Identifying Centre and Radius from an Equation

A circle has equation $(x - 1)^2 + (y + 4)^2 = 36$. Find the centre and radius.

1. $(x - a)^2 + (y - b)^2 = r^2$ — Write the standard form alongside the given equation to compare term by term.
2. $x - a = x - 1 \Rightarrow a = 1$ — The number subtracted from $x$ is $a$. Here $a = 1$.
3. $y - b = y + 4 = y - (-4) \Rightarrow b = -4$ — Rewrite $y + 4$ as $y - (-4)$ to see that $b = -4$. The sign flip is the key step [M1].
4. $r^2 = 36 \Rightarrow r = \sqrt{36} = 6$ — $\sqrt{36} = 6$. State $r = 6$, not $r = 36$ [A1].

Answer: Centre $(1, -4)$, radius $6$

5. Worked Practice: Both Directions

Exam questions test both directions: writing an equation from a description, and reading information from a given equation. The worked examples below cover both so you can see the method applied cleanly each time.

Worked example: Writing the Equation — Fractional and Surd Radius

Write down the equation of the circle with centre $(0, -3)$ and radius $\sqrt{11}$.

1. $(x - a)^2 + (y - b)^2 = r^2$ — Always start here to avoid sign errors.
2. $a = 0, \quad b = -3, \quad r = \sqrt{11}$ — Centre is $(0, -3)$, radius is $\sqrt{11}$.
3. $(x - 0)^2 + (y - (-3))^2 = (\sqrt{11})^2$ — $(\sqrt{11})^2 = 11$. Squaring a square root removes the root sign.
4. $x^2 + (y + 3)^2 = 11$ — $(x - 0)^2 = x^2$. Final answer is clean and correct [A1].

Answer: $x^2 + (y + 3)^2 = 11$

When the radius is a surd such as $\sqrt{11}$, squaring it gives a whole number: $(\sqrt{11})^2 = 11$. The right-hand side of the equation will always be a positive number, never a surd.

7. Exam Technique and Common Mistakes

Two mistakes account for the majority of lost marks on circle equation questions. Knowing them in advance means you can check for them before moving on.

Both specification requirements are now fully covered: you can interpret $(x - a)^2 + (y - b)^2 = r^2$ to extract the centre and radius, and you can write the equation of any circle given its centre and radius.