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Coordinate Geometry of the Circle

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CIE IGCSE Additional Maths

Coordinate Geometry of the Circle

Every circle is just a distance equation in disguise — once you know a circle's centre and radius, you can find where lines touch it, cross it, or miss it completely, using nothing but algebra you already know (Pythagoras, gradients, and simultaneous equations).

Summary — What This Chapter Covers

Equation of a circle

Centre & radius from any form

Line meets circle (0, 1, or 2 times)

Tangents to circles

Two circles intersecting

Common chord equation

A circle with centre (a, b) and radius r has equation (x−a)² + (y−b)² = r². This comes straight from the distance formula.
Circles can also appear in general form x² + y² + 2gx + 2fy + c = 0 — you convert this back to the centre-radius form by completing the square.
To find where a line meets a circle, substitute the line into the circle equation, simplify to a quadratic, and count the solutions: 2 solutions = crosses twice, 1 (repeated) solution = tangent, 0 real solutions = misses entirely.
A tangent is always perpendicular to the radius at the point of contact — this is the single most important fact in this whole chapter.
Two circles can intersect twice, touch once, or not intersect at all — this is decided purely by comparing the distance between centres to the sum and difference of the two radii.
The common chord (the line joining two intersection points of two circles) can be found either by subtracting the circle equations, or from the two intersection points directly.

1. The Equation of a Circle

Where does the formula even come from?

Picture a circle with its centre sat at the point (a, b), and radius r. Now pick any point (x, y) that lies exactly on the circle's edge. By definition, that point is always a distance r away from the centre — no matter where on the circle it sits.

Recall the distance formula between two points: distance = √[(x₂−x₁)² + (y₂−y₁)²]. Applying that between our point (x, y) and the centre (a, b), and setting the distance equal to r, then squaring both sides to remove the square root, gives you the circle equation. It's not a rule to memorise blindly — it's Pythagoras' theorem wearing a disguise.

y ^ | | (x, y) <- any point on circle | /| | / | | r | | / | | *----+ | (a,b) +----------------> x distance from (a,b) to (x,y) is always r => (x-a)^2 + (y-b)^2 = r^2

Core Formula A circle with centre (a, b) and radius r has equation:
(x − a)² + (y − b)² = r²

Reading the equation instantly Whatever number is being subtracted from x inside the bracket is a (the x-coordinate of the centre) — and whatever is subtracted from y is b. The number on the right-hand side is r², not r — always square-root it to get the actual radius.

Practice Question 1

Write down the equation of the circle with centre (−3, 2) and radius 7.

Practice Question 2

A circle has centre (−5, −7) and passes through the point (1, 1). Find the equation of the circle.

Finding the centre and radius from a given equation

This is the reverse process. Given an equation already in the form (x − a)² + (y − b)² = r²:

The radius is the positive square root of whatever number sits on the right-hand side.
The centre coordinates are the values of x and y that would make each bracket equal zero.

The classic sign-flip trap The numbers inside the brackets have the opposite sign to the actual centre coordinates. For example, (x + 12)² + (y − 9)² = 73 has centre (−12, 9) — not (12, −9). Students lose easy marks here constantly. Always ask: "what value of x makes this bracket zero?"

The General Form of a Circle

Sometimes a circle equation is deliberately given in an expanded, messier form: x² + y² + 2gx + 2fy + c = 0. This is called the general form. It's mathematically identical to the centre-radius form — it's just been multiplied out. To get back to something useful, you rearrange it using completing the square on both the x-terms and the y-terms separately.

Worked Example — Completing the Square Find the centre and radius of: x² − 6x + y² + y − 15 = 0

Step 1 — Complete the square on the x-terms:
x² − 6x = (x − 3)² − 9

Step 2 — Complete the square on the y-terms:
y² + y = (y + ½)² − ¼

Step 3 — Substitute back in and simplify:
(x − 3)² − 9 + (y + ½)² − ¼ − 15 = 0
(x − 3)² + (y + ½)² = 9 + ¼ + 15 = 97/4

Centre = (3, −½) Radius = √(97/4) = √97 / 2

Practice Question 3

A circle has the equation x² + 2x + y² − 6y + k = 0 and passes through the point (5, 12). Find the centre and radius of the circle.

2. Intersection of a Circle and a Line

A straight line drawn near a circle can only ever do one of three things to it:

Relationship	What happens	Number of solutions
Crosses through	Line enters and exits the circle — the piece of line inside the circle is called a chord (or a diameter if it passes through the centre)	2 real solutions
Touches (tangent)	Line grazes the circle at exactly one point without crossing into it	1 repeated solution
Misses entirely	Line passes by without ever touching the circle	0 real solutions

The Method: Substitute, Simplify, Solve

4-Step Method Step 1 — Substitute the linear equation (y = mx + c) into the circle equation.
Step 2 — Expand and simplify into a quadratic equation.
Step 3 — Solve the quadratic. Count the number of real solutions to know how many times they intersect.
Step 4 — If asked for coordinates, substitute your x-values (or y-values) back into the line equation to get the matching pair.

Worked example: Does the line y = x − 4 intersect the circle (x − 5)² + (y − 2)² = 13?

Substitute: (x − 5)² + ((x − 4) − 2)² = 13
Expand: (x − 5)² + (x − 6)² = 13 → x² − 10x + 25 + x² − 12x + 36 − 13 = 0
Simplify: x² − 11x + 24 = 0 → (x − 3)(x − 8) = 0 → x = 3 or x = 8
Two solutions → the line crosses the circle twice
Substituting back: y = 3 − 4 = −1, and y = 8 − 4 = 4
Intersection points: (3, −1) and (8, 4)

Shortcut for horizontal/vertical lines You don't need the full method if the line is horizontal or vertical. A horizontal line y = k — just substitute k in for y in the circle equation and solve for x. A vertical line x = k — substitute k in for x and solve for y. Much faster than the general method.

Practice Question 4

Show that the line y = 4x + 4 is tangent to the circle (x − 4)² + (y − 3)² = 17, and find the point of contact.

3. Tangents to Circles

The one fact that unlocks this whole topic

The golden rule A tangent to a circle is always perpendicular to the radius drawn to the point where it touches. This single fact is what lets you find a tangent's equation without ever substituting a line into the circle and solving a quadratic.

TANGENT LINE \ \ \ <- 90° angle here CENTRE *-----• P (point of tangency) (a,b) radius (this segment) Tangent gradient x Radius gradient = -1 (perpendicular)

Finding the equation of a tangent at a known point

Say you know the circle's centre O(x₁, y₁) and a point P(x₂, y₂) that lies on the circle. Here's exactly how to get the tangent line at P:

3-Step Method Step 1 — Find the gradient of the radius OP: m_radius = (y₂ − y₁) / (x₂ − x₁)
Step 2 — Since tangent ⊥ radius, flip and negate it (negative reciprocal): m_tangent = −1 / m_radius
Step 3 — Use y − y₂ = m_tangent(x − x₂) to write the tangent's equation, since it must pass through P.

Faster route If you're comfortable with the algebra, you can skip straight to the tangent gradient using: gradient of tangent = −(x₂ − x₁)/(y₂ − y₁) — this saves you calculating the radius gradient as a separate step.

Worked example: Circle C has equation (x − 5)² + (y − 2)² = 25. Point P(2, 6) lies on C. Find the tangent to C at P.

Step 1 — C has centre (5, 2). Gradient of radius at P = (6 − 2)/(2 − 5) = 4 / (−3) = −4/3
Step 2 — Gradient of tangent = negative reciprocal of −4/3 = 3/4
Step 3 — Line through (2, 6) with gradient 3/4:
y = ¾(x − 2) + 6 = ¾x − 3/2 + 6
Tangent equation: y = ¾x + 9/2

Practice Question 5

A circle has centre (1, −3) and the point (4, 1) lies on the circle. Find the equation of the tangent to the circle at that point.

4. Intersection of Two Circles

Two circles can only relate to each other in a few ways, and — brilliantly — you can work out which one applies without drawing anything, just by comparing the distance between their centres to their two radii.

Step 1: Find the distance between the centres

This is a direct application of Pythagoras' theorem (the same distance formula from Topic 1):

For centres (x₁, y₁) and (x₂, y₂):
d² = (x₂ − x₁)² + (y₂ − y₁)²

Step 2: Compare d to the radii

Let r₁ and r₂ be the two radii, where r₂ ≥ r₁ (i.e. r₂ is the bigger one).

Condition	What it means
`d > r₁ + r₂`	Circles are too far apart — no intersection (0 points)
`d = r₁ + r₂`	Circles just touch on the outside — 1 intersection (external tangent)
`r₂ − r₁ < d < r₁ + r₂`	Circles genuinely overlap — 2 intersections
`d = r₂ − r₁`	Smaller circle sits just inside, touching from within — 1 intersection (internal tangent)
`d < r₂ − r₁`	Smaller circle is completely enclosed inside the bigger one — no intersection

Don't memorise — visualise Instead of memorising these five conditions word-for-word, sketch two circles in your head (or on paper) sliding towards each other. As they get closer they eventually touch (tangent), then overlap (2 points), then one could swallow the other completely, touching again from inside before vanishing entirely. The formulas just describe that story in numbers.

Finding the actual intersection points

5-Step Method Step 1 — Rearrange both circle equations so one side equals zero.
Step 2 — Set the two circle equations equal to each other (this is "solving simultaneously").
Step 3 — Expand and simplify. The x² and y² terms will always cancel out, leaving a simple linear equation.
Step 4 — Substitute this linear equation into either circle equation to solve for the remaining coordinate.
Step 5 — Substitute back to find the matching coordinate pair(s).

Worked example: Find where x² + y² = 2 and (x − 4)² + y² = 10 intersect.

First check they do intersect: centres (0,0) and (4,0), so d = 4.
r₁ = √2 ≈ 1.41, r₂ = √10 ≈ 3.16 → r₁+r₂ ≈ 4.58, r₂−r₁ ≈ 1.75
Since 1.75 < 4 < 4.58, the circles do intersect twice. ✓

Step 1 — x² + y² − 2 = 0 and (x − 4)² + y² − 10 = 0
Step 2 — x² + y² − 2 = (x − 4)² + y² − 10
Step 3 — x² + 8 = x² − 8x + 16 → 8x − 8 = 0 → x = 1
Step 4 — Substitute x = 1 into x² + y² = 2: 1 + y² = 2 → y² = 1 → y = ±1
Intersection points: (1, 1) and (1, −1)

Practice Question 6

Determine whether the circles x² + y² = 4 and (x − 6)² + y² = 4 intersect, touch, or don't meet at all.

5. Equation of a Common Chord

When two circles cross at two points, the straight line joining those two points is called the common chord. It's a chord of both circles simultaneously.

___________ / \ / circle A \ | *----+---+---* intersection point 1 | / \./ \ \ | / | \ | \ / common \ / / \___/___chord____\/__/ \ \ \ circle B \ \____________\ * intersection point 2

Two ways to find it

Method A — If you know both circle equations: subtract one from the other (put both in the form "= 0" and equate them). The squared terms cancel, leaving you with the straight-line equation directly.

Example x² + y² − 4x + 2y − 8 = x² + y² − 8x − 2y + 16
4x + 4y = 24
Common chord: y = 6 − x

Method B — If you know the two intersection points: just use the ordinary straight-line-through-two-points method you already know from earlier coordinate geometry.

Formula For intersection points (x₁, y₁) and (x₂, y₂):
Gradient: m = (y₂ − y₁) / (x₂ − x₁)
Then: y − y₁ = m(x − x₁)

Worked example: Two circles intersect at (3, −1) and (8, 4). Find the common chord.

m = (4 − (−1)) / (8 − 3) = 5/5 = 1
y − (−1) = 1(x − 3)
y + 1 = x − 3
Common chord: y = x − 4

Practice Question 7

Two circles have equations x² + y² − 2x + 4y − 4 = 0 and x² + y² + 6x − 2y − 12 = 0. Find the equation of their common chord.

What to Memorise

Circle equation

(x−a)² + (y−b)² = r² — centre (a,b), radius r. Signs inside brackets are opposite to the actual centre coordinates.

General form

x² + y² + 2gx + 2fy + c = 0 — always convert back via completing the square before reading off centre/radius.

Distance between two points

d = √[(x₂−x₁)² + (y₂−y₁)²] — the basis of everything in this chapter.

Perpendicular gradients

If two lines are perpendicular, m₁ × m₂ = −1. Tangent gradient is always the negative reciprocal of the radius gradient.

Line meets circle

Substitute → simplify to quadratic → count real roots: 2 = crosses, 1 = tangent, 0 = misses.

Two circles

Compare d to r₁+r₂ and r₂−r₁ to know if they intersect twice, touch once, or don't meet — always check r₂ ≥ r₁.

Concepts Checklist

I can write down the equation of a circle given its centre and radius
I can find the centre and radius of a circle given a point that lies on it
I can read off the centre and radius directly from an equation already in (x−a)²+(y−b)²=r² form
I can convert general form (x²+y²+2gx+2fy+c=0) into centre-radius form using completing the square
I can substitute a line into a circle equation and determine how many times they intersect
I understand why a repeated root means the line is a tangent
I know that a tangent is always perpendicular to the radius at the point of contact
I can find the equation of a tangent to a circle at a given point
I can determine whether two circles intersect twice, touch once, or don't meet, using distance between centres
I can find the coordinates where two circles intersect
I can find the equation of the common chord of two intersecting circles (both methods)

Exam Tips & Common Mistakes

Sign errors in the centre The #1 mark-loser in this whole topic. (x + 12)² + (y − 9)² has centre (−12, 9), not (12, −9). Always find the value that makes the bracket zero — don't just "read off" the number you see.

Forgetting to square-root for the radius The right-hand side of the circle equation is r², not r. If the equation ends in "= 49", the radius is 7, not 49. This mistake is especially easy to make under exam time pressure.

Completing the square errors When rearranging general form, students often forget to carry the "−(half-coefficient)²" correction term through to the final constant. Do the x-part and y-part completely separately, then combine everything (including the original constant c) at the very end.

What examiners actually check for

Correct substitution of the linear equation into the circle equation (not the other way round)
Full expansion shown — mark schemes usually award a method mark just for correct expansion
Explicit statement of "1 repeated root → tangent" or "2 roots → intersects twice" — don't just leave the quadratic solved with no conclusion
For tangent questions: showing the perpendicular gradient step clearly, not just stating the final gradient

Quick sanity check habit Whenever you find a centre and radius, sketch a rough circle (even a tiny scribble) and check your line or second circle genuinely looks like it should intersect the way your algebra says. This catches sign errors before you submit your answer.

Also in the full note

Exam Tips & Common Mistakes

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