A circle is just an equation away from being your favourite topic
Here's something that catches most students off guard: coordinate geometry of the circle is one of the most predictable topics on the IGCSE Additional Mathematics paper. The question types repeat. The methods are formulaic. And once you've nailed the three or four core techniques, you can walk into almost any circle question and know exactly what to do. So why does it trip people up? Usually because they rush past the setup and jump straight into algebra without a clear plan.
This guide breaks the topic into the pieces that actually matter for your exam, with worked examples that show you not just the answer but the thinking behind each step.
Two forms of the circle equation (and why you need both)
Every circle question on the IGCSE Additional Mathematics paper revolves around one of two equations. You need to recognise both instantly and switch between them without hesitation.
Standard form
The standard form is the one you can read like a sentence:
(x - a)^2 + (y - b)^2 = r^2
- Centre: (a, b)
- Radius: r
So (x - 3)^2 + (y + 1)^2 = 25 is a circle with centre (3, -1) and radius 5. Notice the sign flip: "y + 1" in the bracket means the y-coordinate of the centre is -1, not +1. That's the single most common misread on circle questions.
General form
The general form looks messier but carries the same information:
x^2 + y^2 + 2gx + 2fy + c = 0
- Centre: (-g, -f)
- Radius: sqrt(g^2 + f^2 - c)
For x^2 + y^2 - 6x + 2y - 15 = 0, you'd read off 2g = -6 so g = -3, and 2f = 2 so f = 1. Centre is (3, -1), and the radius is sqrt(9 + 1 + 15) = sqrt(25) = 5. Same circle as before, just wearing different clothes.
Converting between forms: completing the square
Exam questions love giving you the general form and asking for the centre and radius. The bridge between the two forms is completing the square, and it follows the same steps every single time.
Worked example 1
Find the centre and radius of the circle x^2 + y^2 - 8x + 4y - 5 = 0.
- Group the x terms and y terms: (x^2 - 8x) + (y^2 + 4y) = 5
- Complete the square for x: (x^2 - 8x + 16) - 16, giving (x - 4)^2 - 16
- Complete the square for y: (y^2 + 4y + 4) - 4, giving (y + 2)^2 - 4
- Substitute back: (x - 4)^2 - 16 + (y + 2)^2 - 4 = 5
- Simplify: (x - 4)^2 + (y + 2)^2 = 25
Centre is (4, -2). Radius is 5.
That's it. Every conversion follows those five steps. The numbers change; the method doesn't.
Circle-line intersection: the discriminant decides everything
What happens when a straight line meets a circle? There are exactly three possibilities, and the discriminant of a quadratic tells you which one you're dealing with.
| Discriminant (b^2 - 4ac) | What it means geometrically | Number of intersection points |
|---|---|---|
| Positive (> 0) | The line cuts through the circle | 2 points |
| Zero (= 0) | The line is a tangent to the circle | 1 point (touching) |
| Negative (< 0) | The line misses the circle entirely | 0 points |
The method is always the same: substitute the line equation into the circle equation, rearrange into a quadratic, and check the discriminant.
Worked example 2
Find the points where the line y = x + 1 intersects the circle x^2 + y^2 = 25.
- Substitute y = x + 1 into the circle: x^2 + (x + 1)^2 = 25
- Expand: x^2 + x^2 + 2x + 1 = 25
- Simplify: 2x^2 + 2x - 24 = 0, which gives x^2 + x - 12 = 0
- Factorise: (x + 4)(x - 3) = 0, so x = -4 or x = 3
- Find y values: when x = -4, y = -3; when x = 3, y = 4
The intersection points are (-4, -3) and (3, 4). The discriminant here is 1 + 48 = 49, which is positive, confirming two intersection points before you even solve.
Tangents: the perpendicular trick that works every time
Here's the key insight that makes tangent questions straightforward: a tangent to a circle is always perpendicular to the radius at the point of contact. Always. No exceptions. And perpendicular lines have gradients that multiply to give -1.
So if you know the centre and the point of tangency, you can find the gradient of the radius, flip it to get the gradient of the tangent, and write the equation. No calculus needed.
Worked example 3
Find the equation of the tangent to the circle (x - 2)^2 + (y - 3)^2 = 20 at the point (6, 5).
- Centre is (2, 3). Point of tangency is (6, 5).
- Gradient of radius = (5 - 3)/(6 - 2) = 2/4 = 1/2
- Gradient of tangent = -1 / (1/2) = -2 (negative reciprocal)
- Using y - y1 = m(x - x1): y - 5 = -2(x - 6)
- Simplify: y = -2x + 17
Done. Three pieces of information (centre, point, perpendicularity) give you the tangent equation in five lines of working.
What about tangents from an external point?
Sometimes you're given a point outside the circle and asked for the tangent lines from that point. The approach changes slightly:
- Write the tangent as y - y1 = m(x - x1), keeping m unknown
- Substitute into the circle equation to get a quadratic in x
- Set the discriminant equal to zero (because a tangent touches at exactly one point)
- Solve for m
You'll usually get two values of m, giving two tangent lines. That makes geometric sense: from any external point, you can draw exactly two tangents to a circle.
Intersection of two circles
When two circles intersect, finding their common chord involves a surprisingly clean algebraic trick. Write both circles in general form, then subtract one equation from the other. The x^2 and y^2 terms cancel, leaving you with a linear equation. That linear equation is the common chord.
Worked example 4
Find the equation of the common chord of x^2 + y^2 - 4x + 2y - 11 = 0 and x^2 + y^2 - 2x - 6y + 1 = 0.
Subtract the second from the first:
(-4x + 2y - 11) - (-2x - 6y + 1) = 0
-4x + 2y - 11 + 2x + 6y - 1 = 0
-2x + 8y - 12 = 0
x - 4y + 6 = 0
That's the common chord. If you need the actual intersection points, substitute this line back into either circle equation and solve the resulting quadratic.
The mistakes that cost marks (and how to dodge them)
After working through hundreds of student papers, certain errors come up again and again on circle questions. Here's what to watch for.
| Mistake | Why it happens | How to avoid it |
|---|---|---|
| Sign errors when reading centre from standard form | (x + 3) is read as centre x = 3 instead of x = -3 | Rewrite the bracket as (x - (-3)) to make the sign explicit |
| Forgetting to halve coefficients in general form | Reading 2g = -6 as g = -6 instead of g = -3 | Circle the "2g" and "2f" and physically write the halving step |
| Writing r instead of r^2 in the equation | If radius is 5, writing = 5 instead of = 25 | Always square the radius as a separate step before writing the equation |
| Skipping the discriminant check | Assuming intersection exists without verifying | Calculate the discriminant first; if negative, state "no intersection" and move on |
| Using calculus for tangents | Habit from differentiation topics | The syllabus explicitly states no calculus for circle tangents; use perpendicular gradients |
Pulling it all together: a multi-part question walkthrough
Real IGCSE Additional Mathematics questions often combine several of these techniques in one question. Here's the kind of thing you might face.
A circle has equation x^2 + y^2 + 6x - 10y + 9 = 0.
(a) Find the centre and radius.
(b) Show that the point (0, 1) lies on the circle.
(c) Find the equation of the tangent at (0, 1).
Part (a): Complete the square. (x^2 + 6x + 9) - 9 + (y^2 - 10y + 25) - 25 + 9 = 0, so (x + 3)^2 + (y - 5)^2 = 25. Centre (-3, 5), radius 5.
Part (b): Substitute (0, 1): 0 + 0 + 0 - 10 + 9 = -1. That's not zero, so... wait. Let's be more careful. Substituting into the original: 0 + 1 + 0 - 10 + 9 = 0. Yes, it lies on the circle. (This is exactly the kind of moment where rushing costs marks.)
Part (c): Gradient of radius from (-3, 5) to (0, 1) is (1 - 5)/(0 - (-3)) = -4/3. Tangent gradient = 3/4. Equation: y - 1 = (3/4)(x - 0), giving y = (3/4)x + 1, or equivalently 3x - 4y + 4 = 0.
Self-check questions
Try these without looking at the methods above. If you get stuck, that's your revision target for tonight.
- Write down the equation of a circle with centre (5, -2) and radius 7.
- Find the centre and radius of x^2 + y^2 + 10x - 4y + 13 = 0.
- Determine whether the line y = 2x + 3 intersects the circle x^2 + y^2 = 5. If so, find the points.
- Find the equation of the tangent to x^2 + y^2 = 50 at the point (5, -5).
- Two circles are given by x^2 + y^2 = 25 and x^2 + y^2 - 8x + 6y - 11 = 0. Find the equation of their common chord.
Your revision game plan
Circle questions on the Cambridge IGCSE Additional Mathematics paper are high-reward once you've internalised the core moves. Here's a realistic practice schedule:
- Day 1: Practise converting five circles from general to standard form. Speed matters here.
- Day 2: Work through three circle-line intersection problems. Focus on setting up the substitution cleanly.
- Day 3: Tackle tangent questions, both at a given point and from an external point.
- Day 4: Try two or three full past paper questions that combine multiple parts.
Four days. That's genuinely all it takes to go from shaky to solid on this topic. The methods are mechanical, the algebra is manageable with care, and the question patterns are some of the most predictable on the entire IGCSE Additional Mathematics paper. Once you've done the reps, circle questions become free marks.
A practical guide to coordinate geometry of the circle for IGCSE Additional Mathematics (0606), covering both standard and general forms of the circle equation, completing the square to convert between forms, circle-line intersection problems, tangent equations using the perpendicular-radius property, and intersection of two circles, with worked examples and common pitfalls throughout.
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