When a straight line cuts a circle, it forms a chord. The chord’s length can be found without solving two simultaneous equations for its endpoints: calculate the perpendicular distance d from the circle’s centre to the line, then use the right triangle formed by the radius and half-chord. For radius r, the chord length is 2√(r² − d²), provided 0 ≤ d ≤ r.
For example, the circle (x − 2)² + (y + 1)² = 25 has centre (2, −1) and radius 5. The line 3x + 4y = 12 is two units from the centre, so its chord length is 2√21. This guide derives the formula, verifies the example by coordinates, distinguishes a secant from a tangent and explains how to work backwards from a known chord.
The geometric reason the chord formula works
Draw a circle with centre O and a chord AB. The perpendicular from O to the chord meets the chord at its midpoint M. Therefore AM = MB, and triangle OMA is right-angled.
Let the circle’s radius be r, the perpendicular distance OM be d and the half-chord AM be h. Pythagoras gives h² + d² = r², so h = √(r² − d²). Since AB is twice AM, its length is 2√(r² − d²).
This relationship is the same whether the chord is horizontal, vertical or sloping. The only distance that matters is the shortest distance from the centre to the chord’s supporting line, not a horizontal separation guessed from a drawing.
The key prerequisite is understanding why the perpendicular from the circle’s centre bisects a chord. Once that geometry is clear, the algebra becomes a direct use of the right-triangle theorem.
Read the centre and radius correctly
A circle written (x − h)² + (y − k)² = r² has centre (h, k) and radius r, where r is the positive square root of the number on the right.
For (x − 2)² + (y + 1)² = 25, the centre is O(2, −1), not (−2, 1). The squared radius is 25, so r = 5.
The signs inside the brackets are opposite the corresponding centre coordinates: x − 2 indicates h = 2, while y + 1 is y − (−1), indicating k = −1.
A useful first check is to substitute the proposed centre into the left side and confirm that both squared differences vanish. Reading centre signs incorrectly will corrupt the perpendicular distance and therefore the final chord length.
Rewrite the line in a distance-friendly form
The line 3x + 4y = 12 can be written as 3x + 4y − 12 = 0. In general, for Ax + By + C = 0, the perpendicular distance from the point (h, k) to the line is |Ah + Bk + C|/√(A² + B²).
The absolute value makes distance non-negative. The denominator √(A² + B²) accounts for the orientation and scale of the normal vector (A, B).
For the chosen line, A = 3, B = 4 and C = −12. The denominator is √(9 + 16) = 5.
Do not substitute a line’s gradient in place of A or B without correctly converting its equation. The normal-form coefficients belong to the complete line equation, not just to a coefficient that looks convenient.
Work the complete sloping-line example
For centre O(2, −1) and line 3x + 4y − 12 = 0, calculate d = |3(2) + 4(−1) − 12|/5.
The numerator is |6 − 4 − 12| = |−10| = 10, so d = 10/5 = 2. The radius is 5.
Apply the chord formula: AB = 2√(r² − d²) = 2√(25 − 4) = 2√21.
This is the exact length. It is approximately 9.165 units, which is shorter than the diameter 10, as every non-diameter chord must be. That simple bound provides an immediate reasonableness check.
The main computation took only a centre reading, a point-to-line distance and Pythagoras. No coordinate intersection solving was required.
Why d must not exceed the radius
A straight line meets a circle in two different points exactly when its perpendicular distance d from the centre is strictly less than r.
When d = r, the line touches the circle at one point and is tangent. The chord length formula gives 2√0 = 0, describing a limiting chord that has collapsed to a single contact point.
When d > r, the expression r² − d² is negative. There is no real chord because the line lies entirely outside the circle and has no intersection with the circumference.
This creates a three-case classification: d < r gives a secant line and positive-length chord; d = r gives tangency; d > r gives no circle-line intersection. It prevents students from taking the square root of a negative quantity and inventing a geometric length.
A horizontal chord can be solved almost by inspection
Consider x² + y² = 25 and the horizontal line y = 3. The centre is (0, 0), radius 5 and perpendicular distance to the line 3.
Therefore the chord length is 2√(25 − 9) = 2√16 = 8.
For a direct coordinate check, substitute y = 3 into the circle equation: x² + 9 = 25, so x = −4 or 4. The chord endpoints are (−4, 3) and (4, 3), eight units apart.
Both routes agree. The distance method is efficient, while finding endpoints is useful when the question requests their coordinates as well as the length.
A vertical chord follows the same rule
Keep the circle x² + y² = 25 but use the vertical line x = −3. Its perpendicular distance to the origin is again 3.
The chord length is therefore 8. Directly substituting x = −3 gives 9 + y² = 25, so y = −4 or 4.
The endpoints are (−3, −4) and (−3, 4), also eight units apart. The circle’s symmetry explains why the horizontal and vertical examples produce identical lengths.
The chord formula is orientation-independent. This is helpful in examinations because a sloping line may appear complicated even though the underlying geometry is no different from these two simple examples.
Find the midpoint of the sloping chord
The perpendicular foot from O(h, k) to the line Ax + By + C = 0 is also the chord’s midpoint when the line cuts the circle.
The foot is M = (h, k) − [(Ah + Bk + C)/(A² + B²)](A, B). This formula subtracts the signed component of the centre’s position normal to the line.
In the sloping example, the signed numerator is −10 and A² + B² = 25. Therefore M = (2, −1) − (−10/25)(3, 4) = (2 + 6/5, −1 + 8/5).
Thus M = (16/5, 3/5). Check it lies on the line: 3(16/5) + 4(3/5) = 60/5 = 12. This is the chord’s midpoint, not the circle’s centre.
Recover the exact sloping chord endpoints
For line 3x + 4y = 12, a direction vector parallel to the line is (4, −3), perpendicular to its normal (3, 4). Its length is 5, so a unit direction is (4/5, −3/5).
The half-chord length is √21. Starting from the midpoint M(16/5, 3/5), move √21 units in either direction along the line.
The endpoints are A((16 + 4√21)/5, (3 − 3√21)/5) and B((16 − 4√21)/5, (3 + 3√21)/5).
Their midpoint is M and their difference vector is (8√21/5, −6√21/5), whose length is √[(64 + 36)21/25] = √84 = 2√21. This independent coordinate computation verifies the geometric chord formula.
Why solving simultaneous equations is sometimes longer
You could substitute the line’s equation y = 3 − 3x/4 into the circle (x − 2)² + (y + 1)² = 25 and solve a quadratic in x.
That method is valid, but it introduces fractions, two endpoint coordinates and a further distance calculation. If the only requested quantity is chord length, most of that work is unnecessary.
The geometry-based method produces the length directly from r and d. If a later part asks for the intersection coordinates or chord midpoint, those extra calculations can be added then.
Efficient A-Math problem solving is not simply choosing the quickest shortcut. It is recognising which facts answer the actual question and avoiding work that does not help determine the requested object.
Work backwards when chord length and distance are known
The right triangle gives r² = d² + (L/2)², where L is the full chord length.
Suppose the line of a chord is five units from the circle’s centre and the chord length is 24 units. Half the chord is 12 units, so r² = 5² + 12² = 25 + 144 = 169.
Therefore the radius is 13 units. There is no need to know the chord endpoints.
This reverse version is an important transfer question. Students who only memorise L = 2√(r² − d²) may struggle to rearrange it, while students who understand the right triangle can recover whichever length is missing.
Work backwards from radius and chord length
Suppose a circle has radius 10 and a chord of length 16. Half the chord is 8.
The perpendicular distance d from the centre to the chord is √(10² − 8²) = √(100 − 64) = 6.
Two parallel chords of the same length can lie on opposite sides of the centre, each at distance 6. The calculation gives the magnitude of the separation, not which side the chord occupies.
This distinction matters when reconstructing a line from a centre, direction and chord length. A known distance may describe two parallel candidate lines, not one, unless the problem supplies a side condition.
The longest chord is the diameter
The chord length L = 2√(r² − d²) decreases as the non-negative distance d increases from zero to r.
At d = 0, the line passes through the centre. The chord length is 2r, the diameter, which is the largest possible chord of the circle.
For 0 < d < r, the chord is shorter than the diameter. At d = r, it reduces to a tangent contact point and has zero length as a limiting chord.
This monotonic relationship gives a useful qualitative check: moving a parallel cutting line away from the centre shortens the chord. An answer larger than 2r indicates an incorrect radius, distance or square-root calculation.
Compare two parallel chord lengths
For a circle of radius 5, take two parallel lines at distances d₁ = 0 and d₂ = 3 from its centre.
The central line gives a chord of length 2√25 = 10. The other gives 2√(25 − 9) = 8.
The line closer to the centre creates the longer chord. This is a general fact in one circle: equal distances from the centre produce equal chord lengths, and smaller distances produce longer chords.
The statement must be qualified when comparing different circles. Chords in different circles can have the same length but very different distances from their respective centres because their radii differ. Always identify which circle the geometry refers to.
A tangent distance is a useful limit check
For the circle x² + y² = 25, the horizontal line y = 5 has distance 5 from the centre. It touches at (0, 5).
The chord formula gives 2√(25 − 25) = 0. This signals tangency, not two distinct endpoints separated by zero in an ordinary positive-length chord.
The line y = 6 has distance 6 > 5 and does not meet the circle. The expression under the chord square root would be 25 − 36 = −11, so no real chord length exists.
These simple cases help students understand what the formula’s square-root domain means geometrically. It is not merely a numerical restriction imposed by a calculator.
The point-to-line formula needs consistent coefficients
The distance from (h, k) to Ax + By + C = 0 is |Ah + Bk + C|/√(A² + B²). If a line is given as y = mx + c, rewrite it consistently as mx − y + c = 0 before substituting.
Multiplying the entire line equation by a nonzero number does not change its geometry or the computed distance, because the numerator and denominator scale by the same absolute factor.
For instance, 3x + 4y − 12 = 0 and 6x + 8y − 24 = 0 describe the same line. Using the doubled coefficients gives numerator 20 and denominator 10, still d = 2.
This is a good consistency check when equations appear in different forms. Do not mix A and B from one scaling with C from another.
What if the centre is not provided in centre-radius form?
A circle may be given as x² + y² − 4x + 2y − 20 = 0. Complete the square: (x − 2)² − 4 + (y + 1)² − 1 − 20 = 0.
Rearranging yields (x − 2)² + (y + 1)² = 25. Therefore the centre is (2, −1) and radius 5.
Only after that conversion should you calculate the perpendicular distance to the required chord line.
This makes the complete procedure three stages: find the centre and radius, calculate the centre-to-line distance, then use the right triangle. A student who tries to read radius 20 directly from the general-form constant term is ignoring the contributions from completing the square.
A challenge with an unknown line constant
Consider the circle x² + y² = 25 and horizontal lines y = k. Find values of k for which the intercepted chord is 8 units long.
The centre is (0, 0), radius 5, and perpendicular distance to y = k is |k|. The chord equation gives 8 = 2√(25 − k²).
Divide by two and square: 16 = 25 − k², so k² = 9. Thus k = 3 or k = −3.
There are two valid lines, one above and one below the centre. Both produce an eight-unit chord. The absolute distance |k| explains why two values appear, and substituting either into the chord formula confirms the result.
What a radius and distance can never tell us alone
The pair (r, d) determines a chord’s length but not its orientation or exact position in the coordinate plane.
For the circle x² + y² = 25, both lines y = 3 and x = 3 have centre distance 3 and chord length 8, yet their endpoints differ. Infinitely many directions can produce a chord of that same length.
Therefore, if a problem asks for a chord equation or endpoint coordinates, additional information about its line or orientation is necessary.
This is an information-sufficiency lesson. A mathematical result can be unique for one requested quantity, such as length, while other features of the geometric object remain undetermined.
Common mistakes and first-error repairs
One student reads the centre with the wrong bracket signs. Another reads r² as r. A third calculates vertical separation for a sloping line instead of perpendicular distance. A fourth forgets the factor 2 after finding the half-chord.
These errors need different explanations. Use a centre-substitution check for sign reading, a radius-versus-squared-radius comparison for the second problem, and a perpendicular diagram for distance.
Students who correctly find √21 but present it as the full chord have understood Pythagoras but forgotten that the perpendicular bisects the chord. One labelled right triangle can repair that error.
A changed example using a differently oriented line then tests whether the understanding transfers instead of depending on the numbers in the model solution.
Six independent questions with checked answers
Question one: centre (2, −1), radius 5, line 3x + 4y = 12. Chord length: 2√21.
Question two: circle x² + y² = 25, line y = 3. Chord length: 8.
Question three: circle x² + y² = 25, line y = 5. One tangent contact point; limiting chord length zero.
Question four: circle x² + y² = 25, line y = 6. No real chord.
Question five: a chord length of 24 lies five units from its centre. Radius: 13.
Question six: radius 5 and chord length 8 for a horizontal line y = k in x² + y² = 25. Answer: k = −3 or 3.
For each task, write the centre, radius, perpendicular distance and the equation r² = d² + (L/2)² before calculating.
Syllabus suitability and further reading
This problem joins circle equations, distance from a point to a line, coordinate geometry and the geometry of right triangles. The more advanced endpoint-vector method is enrichment where it extends beyond the student’s taught syllabus.
For Singapore, check the relevant SEAB references: 2026 GCE O-Level Additional Mathematics 4049, 2027 SEC G3 K341, or 2027 SEC G2 K232, as applicable.
Related guides include How to Reconstruct a Circle Through Three Points, How to Find a Circle Equation from Diameter Endpoints and How to Find a Circle Tangent to Two Parallel Lines.
Parents can browse the Additional Mathematics archive or contact eduKate Singapore about suitable small-group tutorials near Sixth Avenue MRT. The objective is to recognise the perpendicular right triangle and use the exact information a question supplies. Properly taught kids shine a bright light into the future.