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HomeCIE IGCSE MathematicsCoordinate Geometry: Equation of a circle centred at the origin
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Coordinate Geometry: Equation of a circle centred at the origin

2,034 words · Last updated May 2026

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What you'll learn

This topic covers the equation of a circle with its centre at the origin (0, 0) in coordinate geometry, a fundamental concept in CIE IGCSE Mathematics. You'll learn to derive and apply the equation x² + y² = r², determine whether points lie on, inside or outside a circle, and solve problems involving circles and straight lines. This topic regularly appears in Paper 2 and Paper 4 examinations, often combined with other coordinate geometry concepts.

Key terms and definitions

Circle — the set of all points in a plane that are equidistant from a fixed point called the centre.

Radius — the constant distance from the centre of a circle to any point on its circumference, denoted by r where r > 0.

Origin — the point (0, 0) where the x-axis and y-axis intersect on a coordinate plane.

Equation of a circle centred at the origin — the algebraic relationship x² + y² = r² that describes all points (x, y) lying on a circle of radius r with centre at (0, 0).

Circumference — the complete boundary or perimeter of a circle.

Chord — a straight line segment whose endpoints both lie on the circle.

Tangent — a straight line that touches the circle at exactly one point, perpendicular to the radius at that point.

Pythagorean theorem — the mathematical relationship a² + b² = c² in a right-angled triangle, which forms the foundation for deriving the circle equation.

Core concepts

Deriving the equation x² + y² = r²

The equation of a circle centred at the origin stems directly from Pythagoras' theorem. Consider any point P(x, y) on a circle with centre O(0, 0) and radius r.

The distance from O to P must equal r. Using the distance formula or creating a right-angled triangle:

  • The horizontal distance from O to P is x
  • The vertical distance from O to P is y
  • The hypotenuse (the radius) is r

Applying Pythagoras' theorem: x² + y² = r²

This equation holds true for every point on the circle. The standard form x² + y² = r² is the equation you must recognise and use in examinations.

Key observations:

  • Both x and y are squared, making the equation symmetrical about both axes
  • The coefficient of x² and y² is always 1
  • There are no x or y terms (no terms with power 1)
  • The right-hand side is always the square of the radius

Finding the radius from the equation

Given an equation in the form x² + y² = k, where k is a positive constant, the radius can be found immediately:

r² = k, therefore r = √k

For example:

  • x² + y² = 25 represents a circle with radius 5 (since √25 = 5)
  • x² + y² = 12 represents a circle with radius √12 or 2√3
  • x² + y² = 0.64 represents a circle with radius 0.8

Important: Always take the positive square root, as radius must be positive. Leave answers in surd form unless the question specifies otherwise or requires a decimal approximation.

Writing the equation given the radius

When given the radius, substitute r² into the standard form.

Process:

  1. Square the radius value
  2. Write x² + y² = [squared radius value]

Examples:

  • Radius = 7 gives x² + y² = 49
  • Radius = 3.5 gives x² + y² = 12.25
  • Radius = √10 gives x² + y² = 10

Determining if a point lies on, inside or outside the circle

For a circle with equation x² + y² = r² and a point (a, b):

Substitution method:

  1. Calculate a² + b²
  2. Compare with r²

Three cases arise:

  • If a² + b² = r², the point lies on the circle
  • If a² + b² < r², the point lies inside the circle
  • If a² + b² > r², the point lies outside the circle

This is a common exam question requiring clear working. Always show the substitution step explicitly.

Example: For the circle x² + y² = 41, test point (4, -5):

  • Substitute: 4² + (-5)² = 16 + 25 = 41
  • Since 41 = 41, the point lies on the circle

Example: For the circle x² + y² = 50, test point (3, 6):

  • Substitute: 3² + 6² = 9 + 36 = 45
  • Since 45 < 50, the point lies inside the circle

Intersection of a circle and a straight line

When a straight line and circle are given, you may need to find their points of intersection. This involves simultaneous equations.

Method:

  1. Write the equation of the line (usually y = mx + c or x = k or y = k)
  2. Write the equation of the circle x² + y² = r²
  3. Substitute the linear equation into the circle equation
  4. Solve the resulting quadratic equation
  5. Find corresponding coordinates

The number of intersection points indicates:

  • Two distinct solutions: line is a secant (cuts through the circle)
  • One solution (repeated root): line is a tangent (touches once)
  • No real solutions: line does not intersect the circle

Finding the length of a chord

A chord is a line segment connecting two points on the circle. To find its length:

  1. Find the coordinates of both endpoints (often by solving simultaneous equations)
  2. Apply the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Alternatively, if you know the perpendicular distance from the centre to the chord and the radius, use the relationship: (half-chord)² + (perpendicular distance)² = r²

Circle and axes intersections

Circles centred at the origin intersect the coordinate axes at specific points that are straightforward to find.

x-axis intersections (where y = 0): Substitute y = 0 into x² + y² = r² x² = r² x = ±r Points: (r, 0) and (-r, 0)

y-axis intersections (where x = 0): Substitute x = 0 into x² + y² = r² y² = r² y = ±r Points: (0, r) and (0, -r)

The circle always crosses each axis at two points, symmetrically placed about the origin, at distance r from the origin.

Worked examples

Example 1: Basic equation and point testing

Question: A circle has its centre at the origin and passes through the point (5, 12).

(a) Find the equation of the circle. [2 marks] (b) Determine whether the point (8, 9) lies inside, on or outside the circle. [2 marks]

Solution:

(a) First, find the radius using the distance formula or Pythagoras: r² = 5² + 12² = 25 + 144 = 169

The equation is x² + y² = 169

Alternative: r = √169 = 13, so x² + y² = 13² = 169

(b) Substitute (8, 9) into the left side: 8² + 9² = 64 + 81 = 145

Compare: 145 < 169

The point (8, 9) lies inside the circle


Example 2: Intersection with a line

Question: A circle has equation x² + y² = 25. The line y = x - 1 intersects the circle.

(a) Find the coordinates of the points of intersection. [4 marks] (b) Calculate the length of the chord formed. [2 marks]

Solution:

(a) Substitute y = x - 1 into x² + y² = 25:

x² + (x - 1)² = 25

Expand: x² + x² - 2x + 1 = 25

Simplify: 2x² - 2x + 1 = 25

2x² - 2x - 24 = 0

Divide by 2: x² - x - 12 = 0

Factorise: (x - 4)(x + 3) = 0

Therefore: x = 4 or x = -3

When x = 4: y = 4 - 1 = 3, giving point (4, 3)

When x = -3: y = -3 - 1 = -4, giving point (-3, -4)

(b) Distance between (4, 3) and (-3, -4):

d = √[(4 - (-3))² + (3 - (-4))²]

d = √[7² + 7²]

d = √[49 + 49]

d = √98 = 7√2 (or 9.90 to 3 s.f.)


Example 3: Finding radius in surd form

Question: A circle centred at the origin has equation x² + y² = 20.

(a) Write down the exact value of the radius in its simplest form. [2 marks] (b) The point (2, k) lies on the circle, where k > 0. Find the value of k. [2 marks]

Solution:

(a) r² = 20

r = √20 = √(4 × 5) = 2√5

(b) Substitute (2, k) into x² + y² = 20:

2² + k² = 20

4 + k² = 20

k² = 16

k = 4 (since k > 0)

Therefore k = 4

Common mistakes and how to avoid them

Mistake: Writing the equation as x² + y² = r instead of x² + y² = r² Correction: Always square the radius. The equation contains r-squared, not r. If radius = 6, write x² + y² = 36, not x² + y² = 6.

Mistake: Forgetting to square negative coordinates when substituting a point Correction: Remember (-5)² = 25, not -25. When testing point (-3, 4), calculate (-3)² + 4² = 9 + 16 = 25, showing all working clearly.

Mistake: Incorrectly simplifying surds for the radius Correction: If x² + y² = 48, the radius is √48 = √(16 × 3) = 4√3, not 4√12 or left as √48 when simplification is possible.

Mistake: Stating a point is "not on the circle" rather than specifying inside or outside Correction: When a² + b² ≠ r², you must determine whether the point is inside (a² + b² < r²) or outside (a² + b² > r²). The question specifically asks which.

Mistake: Finding only one intersection point when solving simultaneous equations Correction: Circle-line intersections typically yield a quadratic equation with two solutions. Find both x-values and their corresponding y-values. Check your factorisation or use the quadratic formula.

Mistake: Taking the negative square root for radius Correction: Radius is always positive by definition. If x² + y² = 64, then r = 8, never r = -8. Only use the positive root.

Exam technique for "Coordinate Geometry: Equation of a circle centred at the origin"

Command word "Find the equation" requires you to write x² + y² = [specific number]. You must calculate r² by finding the distance from the origin to a given point, or by squaring a stated radius. Show working for the calculation of r² to earn method marks even if your final answer contains an error.

Command word "Show that" means the answer is provided—your task is to demonstrate the result through clear algebraic steps. When showing a point lies on a circle, substitute both coordinates, calculate the sum of squares, and explicitly state this equals r². Write a conclusion sentence such as "Since LHS = RHS, the point lies on the circle."

Simultaneous equations questions typically award 1 mark for correct substitution, 2 marks for solving the quadratic, and 1 mark for finding all coordinate pairs. Set out your work clearly: substitute, expand, collect terms, solve, then find corresponding values. Check both solutions.

Accuracy requirements: Unless stated otherwise, leave answers involving surds in exact form (e.g., √18 = 3√2). For decimal approximations, use at least 3 significant figures unless specified. Mark schemes penalise premature rounding, so keep full calculator values in intermediate steps.

Quick revision summary

The equation of a circle centred at the origin (0, 0) with radius r is x² + y² = r². To find r from the equation, take the positive square root of the right-hand side. Test whether point (a, b) lies on the circle by checking if a² + b² = r²; if less than r², it's inside; if greater, it's outside. To find intersection points with a line, substitute the line equation into the circle equation and solve the resulting quadratic. Common errors include forgetting to square r and mishandling negative coordinates. This topic combines algebra and coordinate geometry skills essential for IGCSE examination success.

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