Circle Equations and Properties Revision Notes | GCSE AQA Higher Maths

1. Why Circle Equations Matter

Picture a radar screen: every blip at the same distance from the centre sits on a perfect circle. That distance relationship is exactly what the equation of a circle captures — and once you see it, the formula writes itself.

At GCSE Higher level, the examiner expects you to recognise, use, and sketch the equation of a circle whose centre sits at the origin. This lesson builds that skill from the ground up, then sharpens it into exam technique.

2. Building the Equation from Pythagoras

Here is the key insight: a circle is the set of all points that are the same distance from one fixed point. When that fixed point is the origin $(0, 0)$, you already have the tool to write the equation — Pythagoras' theorem.

Take any point $P = (x, y)$ on the circle. Drop a right-angled triangle from $P$ to the $x$-axis and back to the origin. The horizontal leg has length $x$, the vertical leg has length $y$, and the hypotenuse is the radius $r$. Pythagoras gives $x^2 + y^2 = r^2$ — and that is the equation of the circle.

Two things to fix in your memory: the centre is always $(0, 0)$ for this form, and the right-hand side is $r^2$, not $r$. Examiners regularly present the equation with a non-square number on the right — your first job is always to square-root it to find the actual radius.

3. Reading and Sketching Circle Equations

Once you have the radius, sketching is straightforward. A circle centred at the origin always passes through four key points: $(r, 0)$, $(-r, 0)$, $(0, r)$, and $(0, -r)$. Mark these, then draw a smooth curve through them.

Worked example: Sketching a Circle from its Equation

Sketch the circle with equation $x^2 + y^2 = 25$, labelling the centre and the points where the circle crosses the axes. (calculator-allowed exam question approach)

1. The equation $x^2 + y^2 = 25$ is already in the form $x^2 + y^2 = r^2$, so $r^2 = 25$. — Recognise the standard form immediately [M1].
2. $r = \sqrt{25} = 5$ — Square-root the right-hand side. $5 \times 5 = 25$ confirms this is exact.
3. Centre $= (0, 0)$ — For $x^2 + y^2 = r^2$, the centre is always the origin.
4. $(5, 0),\ (-5, 0),\ (0, 5),\ (0, -5)$ — Substitute $y = 0$ to get $x = \pm 5$; substitute $x = 0$ to get $y = \pm 5$ [A1].
5. Draw a smooth circle of radius 5 centred at the origin, passing through all four intercepts. — Label the centre O and mark all four intercept coordinates. A sketch does not need to be exact — key features labelled earns full marks.

Answer: Circle centred at $(0, 0)$ with radius $5$, crossing the axes at $(\pm 5, 0)$ and $(0, \pm 5)$.

Notice the circle is perfectly symmetrical about both axes and the origin. This symmetry is a direct consequence of the equation: replacing $x$ with $-x$ or $y$ with $-y$ leaves $x^2 + y^2 = r^2$ unchanged.

4. Points On, Inside, and Outside a Circle

The equation $x^2 + y^2 = r^2$ is a test you can apply to any coordinate pair. Substitute the $x$ and $y$ values, compute $x^2 + y^2$, and compare the result to $r^2$. The comparison tells you exactly where the point sits relative to the circle.

Worked example: Determining Whether a Point Lies On, Inside, or Outside a Circle

A circle has equation $x^2 + y^2 = 100$. Determine whether each point lies on, inside, or outside the circle: (a) $(6, 8)$, (b) $(5, 9)$, (c) $(4, 6)$. (Without-a-calculator approach — show all arithmetic.)

1. $r^2 = 100$ — This is our benchmark. Every test compares $x^2 + y^2$ against 100.
2. $6^2 + 8^2 = (6 \times 6) + (8 \times 8) = 36 + 64 = 100$ — $100 = r^2$, so $(6, 8)$ lies ON the circle [A1].
3. $5^2 + 9^2 = (5 \times 5) + (9 \times 9) = 25 + 81 = 106$ — $106 > 100 = r^2$, so $(5, 9)$ lies OUTSIDE the circle [A1].
4. $4^2 + 6^2 = (4 \times 4) + (6 \times 6) = 16 + 36 = 52$ — $52 < 100 = r^2$, so $(4, 6)$ lies INSIDE the circle [A1].

Answer: (a) On the circle. (b) Outside the circle. (c) Inside the circle.

5. Writing the Equation from Given Information

Examiners also work in reverse: they give you the radius (or a point on the circle) and ask you to write the equation. The process is the same formula, used differently — square the radius and place it on the right-hand side.

Worked example: Writing the Equation of a Circle Given a Point on It

A circle is centred at the origin and passes through the point $(3, 4)$. Write the equation of the circle. [2 marks]

1. $r^2 = 3^2 + 4^2 = (3 \times 3) + (4 \times 4) = 9 + 16 = 25$ — Because $(3, 4)$ is on the circle, substituting into $x^2 + y^2 = r^2$ gives $r^2$ directly [M1].
2. $x^2 + y^2 = 25$ — Substitute $r^2 = 25$ into the standard form [A1]. Do not simplify further — this is already the required form.

Answer: $x^2 + y^2 = 25$

You will notice this is the same circle from the sketching example earlier — the point $(3, 4)$ is one of the infinitely many points on that circle. This is a useful self-check: if you sketch the circle and your given point does not appear to sit on the circumference, revisit your arithmetic. In this case, the calculations show that the radius is indeed $r = 5$ derived from the distance formula, confirming it is accurate.

7. Exam Technique and Common Mistakes

The circle equation topic is compact, which means examiners squeeze marks out of precision. The mistakes below are the ones that cost students marks on otherwise correct working — knowing them in advance is a genuine advantage.

One more examiner habit worth knowing: questions sometimes present the equation with the constant on the left, such as $169 = x^2 + y^2$. This is identical to $x^2 + y^2 = 169$ — do not let the rearrangement throw you. The radius is still $\sqrt{169} = 13$.

Higher-tier questions on this topic often combine two or three skills in one question: reading the equation, finding the radius, and testing a point. The worked example below mirrors that style.

8. Pulling It Together — Exam-Style Question

Worked example: Full Exam-Style Question on Circle Equations

A circle $C$ has equation $x^2 + y^2 = 85$. (a) Write down the radius of $C$. (b) Show that the point $P(2, 9)$ lies on $C$. (c) Determine whether $Q(7, 6)$ lies on, inside, or outside $C$, giving a reason. [4 marks total]

1. $r^2 = 85$, so $r = \sqrt{85}$ — Leave in surd form — 85 is not a perfect square. On a calculator-allowed exam question you could write $r \approx 9.22$, but $\sqrt{85}$ is the exact answer [A1].
2. $2^2 + 9^2 = (2 \times 2) + (9 \times 9) = 4 + 81 = 85$ — The result equals $r^2 = 85$, confirming $P$ satisfies the equation [M1].
3. $2^2 + 9^2 = 85 = r^2$, therefore $P(2, 9)$ lies on $C$. — For a 'show that' question, always end with a clear concluding statement linking back to the target [A1].
4. $7^2 + 6^2 = (7 \times 7) + (6 \times 6) = 49 + 36 = 85$ — Wait — $85 = r^2$, so $Q$ also lies on the circle. Let us re-read: $Q = (7, 6)$. $49 + 36 = 85$. $Q$ is ON the circle, not inside or outside. State this clearly [A1].

Answer: (a) $r = \sqrt{85}$. (b) $2^2 + 9^2 = 85 = r^2$, so $P$ lies on $C$. (c) $7^2 + 6^2 = 49 + 36 = 85 = r^2$, so $Q$ also lies on $C$ (not inside or outside).

That last result is a reminder to trust your arithmetic over your assumptions. The question implied $Q$ would be inside or outside, but the calculation showed it is on the circle. In an exam, state what the maths tells you — examiners reward correct reasoning, not the expected answer.