When two circles intersect, their common chord joins the two intersection points. Rather than solving two full quadratic equations simultaneously, subtract one circle equation from the other. The x² and y² terms cancel, leaving the straight line through the common points. This line is often called the common-chord line or radical axis.
For example, x² + y² = 25 and (x − 4)² + y² = 25 intersect where x = 2. Substituting gives y = ±√21, so the common chord has endpoints (2, √21) and (2, −√21) and length 2√21. This guide explains why subtraction works, how to handle unequal radii, how to detect tangency or no intersection, and what information remains when the circles are concentric.
Why subtracting circle equations makes the problem easier
A circle in Cartesian coordinates contains x² and y² with equal coefficients. When two circle equations are written in standard expanded form, the squared terms have the same structure.
Subtracting the equations eliminates x² and y², leaving an equation that is linear in x and y when the two centres are different. Every point that belongs to both circles must satisfy that linear equation.
This transforms a pair of quadratic equations into one straight-line condition plus one remaining circle equation. Finding the common points becomes simpler because the straight line restricts their possible locations.
The subtraction does not automatically prove that real intersections exist. It identifies a candidate line; a second step must check where that line actually meets the circle. This distinction is particularly important when the circles are too far apart to meet.
Read the centres and radii before calculating
Consider the circles C₁: x² + y² = 25 and C₂: (x − 4)² + y² = 25.
C₁ has centre O₁(0, 0) and radius 5. C₂ has centre O₂(4, 0) and the same radius 5. Their centres are four units apart.
Because the distance between the centres is smaller than the sum of the radii, 10, and greater than their radius difference, zero, the circles meet at two points.
A quick sketch shows their symmetry about the x-axis. We should expect one intersection above that axis and another below, both sharing the same x-coordinate. The algebra will determine their exact coordinates and the chord’s length.
Expand only one equation before subtraction
Expand the second circle: (x − 4)² + y² = 25 becomes x² − 8x + 16 + y² = 25.
The first circle is x² + y² = 25. Subtract the first equation from the second. The x² terms cancel, the y² terms cancel and the constants 25 cancel.
What remains is −8x + 16 = 0, so x = 2. Every common point must lie on the vertical line x = 2.
This is the equation of the common-chord line. There is no need to expand a general quadratic in y or apply the quadratic formula merely to determine the line on which the intersection points lie.
Find both common intersection coordinates
Substitute x = 2 into C₁: x² + y² = 25. This gives 4 + y² = 25, so y² = 21.
Therefore y = √21 or y = −√21. The intersection points are A(2, √21) and B(2, −√21).
Both points also satisfy C₂, because its squared horizontal distance is (2 − 4)² = 4, and 4 + 21 = 25.
Keep both square-root signs. A chord has two endpoints when the circles intersect at two distinct points. Reporting only the positive y-value would describe one intersection but not the complete common chord.
Calculate the common-chord length
The two endpoints A and B have the same x-coordinate 2. Their vertical separation is √21 − (−√21) = 2√21.
Thus the common chord has length L = 2√21 units. It is shorter than the diameter 10, as expected because the chord line x = 2 does not pass through either centre.
The midpoint of the chord is ((2 + 2)/2, (√21 − √21)/2) = (2, 0).
The perpendicular distance from O₁(0, 0) to x = 2 is 2. Pythagoras gives half-chord length √(5² − 2²) = √21, confirming the result independently.
The common chord is perpendicular to the line of centres
The centres O₁(0, 0) and O₂(4, 0) lie on the horizontal x-axis. Their joining line is horizontal.
The common-chord line x = 2 is vertical, so it is perpendicular to the line of centres. Its midpoint (2, 0) lies on that centre line.
This is not a coincidence limited to equal circles. For two non-concentric circles with two real intersection points, the straight line through those points is perpendicular to the line joining the centres.
The geometric reason is that each centre has equal distances to both chord endpoints. Therefore both centres lie on the perpendicular bisector of the common chord, and the line through the centres is that perpendicular bisector.
An equal-radius pair has a midpoint chord line
When two circles have the same radius, their common-chord line lies halfway between their centres along the joining direction.
For centres (0, 0) and (4, 0), that halfway position is x = 2. Both intersection points have x-coordinate 2.
But the perpendicular bisector is a line, not necessarily the chord segment itself. The actual chord endpoints depend on the radius and the remaining circle equation.
If the centre separation increased while the radii stayed equal, the chord line would move to the new midpoint, but its length would change. The shared radius fixes how far the chord extends on each side of that midpoint.
A second equal-radius example with vertical centres
Take C₁: x² + y² = 25 and C₂: x² + (y − 6)² = 25. The centres are (0, 0) and (0, 6), six units apart.
Expand and subtract: x² + y² − 12y + 36 = 25 and x² + y² = 25 together give −12y + 36 = 0, so y = 3.
Substituting into the first circle gives x² + 9 = 25, hence x = ±4. The common chord endpoints are (−4, 3) and (4, 3).
Its length is 8. The chord is horizontal because the line of centres is vertical. This changed orientation confirms that the subtraction method is not restricted to centres aligned with the x-axis.
Unequal radii shift the common-chord line
Now let C₁ be x² + y² = 25, with radius 5, and let C₂ be (x − 4)² + y² = 16, with radius 4.
Expand C₂: x² − 8x + 16 + y² = 16, so x² + y² − 8x = 0.
At a common point, C₁ gives x² + y² = 25. Substitute that into the second equation: 25 − 8x = 0, so x = 25/8.
The chord line is no longer the midpoint x = 2. Different radii change the positions of points that belong to both circles, although the common-chord line remains perpendicular to the horizontal line of centres.
Find the unequal-radius chord exactly
With x = 25/8, substitute into x² + y² = 25. We obtain y² = 25 − 625/64 = (1600 − 625)/64 = 975/64.
Since 975 = 25 × 39, this gives y = ±5√39/8. The common points are (25/8, 5√39/8) and (25/8, −5√39/8).
The chord length is twice the positive y-coordinate: L = 5√39/4, approximately 7.806.
This is smaller than the smaller circle’s diameter, which is 8. The bound provides a useful check. A chord common to both circles cannot be longer than either circle’s diameter.
Verify the points satisfy the smaller circle
For the unequal-radius example, the second centre is (4, 0) and its squared radius is 16.
At x = 25/8, the horizontal displacement from the second centre is 25/8 − 4 = −7/8, with square 49/64.
The squared vertical coordinate is 975/64. Their sum is (49 + 975)/64 = 1024/64 = 16.
Therefore each proposed endpoint lies on the second circle as well as the first. This exact verification is stronger than estimating the intersection positions from a sketch, especially when the coordinates involve surds and fractions.
General formula when centres lie on one horizontal line
Suppose the first circle has centre (0, 0) and radius r₁. The second has centre (d, 0) and radius r₂, where d > 0.
Their equations are x² + y² = r₁² and (x − d)² + y² = r₂². Expand the second and subtract the first.
The result is −2dx + d² = r₂² − r₁², so x = (r₁² − r₂² + d²)/(2d).
Call this x-coordinate a. If the circles meet at two points, their y-coordinates are ±√(r₁² − a²), and the common-chord length is 2√(r₁² − a²). This formula applies only when the square root is real; it also assumes distinct centres because d appears in the denominator.
The common-chord line is sometimes called the radical axis
The radical axis of two non-concentric circles is the line on which the squared-distance expressions to their centres, after subtracting the corresponding squared radii, are equal.
At an actual common point, both circle equations equal zero in their expanded forms, so the point lies on that radical axis. If the circles intersect twice, the radical axis is exactly the line through both intersection points.
The term comes from a wider geometric concept known as power of a point. It is useful enrichment because the algebraic subtraction method already gives the same linear equation.
Be precise: the radical axis can exist even when two circles do not intersect in real points. In that case it is a line of equal powers, not a physical chord of the circles.
A line from subtraction does not guarantee a chord
Consider C₁: x² + y² = 25 and C₂: (x − 8)² + y² = 1. Their centres are eight units apart, while their radii add to only six.
Subtracting the equations gives the candidate line x = 11/2. But substituting x = 11/2 into the first circle gives y² = 25 − 121/4 = −21/4.
There is no real y satisfying that equation. The circles are disjoint and have no common chord.
The conclusion must be “no real intersection”, not a chord with an imaginary length. This example is a useful safeguard against treating every linear equation obtained by subtraction as evidence of two actual crossing points.
External tangency gives one common point
Let C₁ be x² + y² = 25 and C₂ be (x − 8)² + y² = 9. The centres are eight units apart, and the radii are 5 and 3.
Their radius sum is also 8, so the circles touch externally at one point. Subtraction gives the common-point line x = (25 − 9 + 64)/(16) = 5.
Substitute x = 5 into C₁: 25 + y² = 25, so y = 0. The only common point is (5, 0).
The common chord has collapsed to one point, giving zero chord length. This is tangency rather than a positive-length chord joining two distinct intersections.
Internal tangency also produces one point
Suppose C₁ has centre (0, 0) and radius 5, while C₂ has centre (3, 0) and radius 2. The distance between centres is three, exactly the difference 5 − 2.
The smaller circle lies inside the larger one and touches it internally at (5, 0). In equations, C₁ is x² + y² = 25 and C₂ is (x − 3)² + y² = 4.
Subtracting gives x = (25 − 4 + 9)/(6) = 5. Substitution into C₁ again yields y = 0.
This illustrates the two kinds of tangency: external when centre distance equals the sum of radii, and internal when it equals the absolute radius difference.
The geometric intersection conditions
For two circles with radii r₁ and r₂ and centre separation d > 0, two distinct common points exist when |r₁ − r₂| < d < r₁ + r₂.
A single tangent contact occurs when d equals r₁ + r₂ or |r₁ − r₂|, subject to the circles being genuinely distinct. If d > r₁ + r₂, the circles are too far apart.
If d < |r₁ − r₂|, one circle lies strictly inside the other without touching. The subtraction line may be mathematically defined but gives no real chord.
Checking these inequalities before lengthy substitution predicts whether the final y² should be positive, zero or negative. It is a geometry-based consistency check for the algebra.
Concentric circles need separate treatment
Two circles are concentric when their centres coincide. If their radii differ, their circumferences never meet, so there is no common chord.
If their centres and radii both coincide, they represent the same circle and share infinitely many circumference points. There is no unique common chord line determined by subtracting the equations.
In the horizontal-centre formula, the denominator 2d becomes zero when d = 0. That is not an invitation to divide by zero; it signals that the distinct-centre derivation no longer applies.
Always inspect whether the centres are different before using a radical-axis formula. Treat coincident circles and concentric unequal-radius circles as exceptional cases.
General centre coordinates still give a straight line
Let the circles have centres (h₁, k₁) and (h₂, k₂), with squared radii r₁² and r₂². Their equations are (x − h₁)² + (y − k₁)² = r₁² and (x − h₂)² + (y − k₂)² = r₂².
Subtract the expanded equations. The x² and y² terms cancel, leaving 2(h₂ − h₁)x + 2(k₂ − k₁)y = r₁² − r₂² + h₂² + k₂² − h₁² − k₁².
That is a straight line when the centres differ. It is perpendicular to the vector connecting the two centres because its normal vector is proportional to (h₂ − h₁, k₂ − k₁).
Use the resulting linear equation with either original circle equation to find the actual intersection coordinates and verify whether real points exist.
Do not solve two full quadratics unless required
A direct solution can substitute the expression for y from one circle equation into another, but this often introduces square roots or complicated algebra.
Subtracting first takes advantage of the shared x² + y² structure and gives a linear relation immediately. That relation usually makes the remaining substitution much simpler.
If the problem asks only for the line of the common chord, the subtraction result may already be the full answer. If it asks for chord length, you must find the distance from the centre to that line or the actual intersection coordinates.
Method efficiency follows the requested quantity. Avoid continuing to calculate coordinates when the question asks only for the linear chord equation.
Find chord length from centre-to-line distance
After obtaining the common-chord line, calculate its perpendicular distance d₁ from the first circle’s centre. If the first radius is r₁ and the line intersects it, the chord length is L = 2√(r₁² − d₁²).
For the equal-radius example, the line is x = 2, the first centre is (0, 0), d₁ = 2 and r₁ = 5. Hence L = 2√(25 − 4) = 2√21.
For the unequal-radius example, d₁ = 25/8, so L = 2√(25 − 625/64) = 5√39/4.
This is an efficient route when the line has been found but individual endpoint coordinates are not required. It connects the two-circle intersection method with ordinary circle chord geometry.
The common chord has one length in both circles
The same segment AB lies on the circumference of both circles. Therefore calculating its length from the first centre and radius must agree with calculating it from the second.
In the unequal-radius example, the chord line x = 25/8 is 25/8 units from O₁(0, 0). Its distance from O₂(4, 0) is |25/8 − 4| = 7/8.
Using the second circle’s radius 4 gives L = 2√[16 − (7/8)²] = 2√[(1024 − 49)/64] = 5√39/4.
This matches the first-circle calculation. Two independent chord-distance checks are especially useful for catching a sign or radius error in the radical-axis calculation.
Common mistakes and the first correction to make
One student subtracts the circle equations but forgets to expand (x − 4)² correctly. Another finds the chord line and assumes it proves intersection without checking the remaining circle equation.
A third finds y² = 21 but reports only y = √21, losing one endpoint. A fourth calculates half the chord length and forgets to multiply by two.
The appropriate repair depends on the first invalid step. Practise expanding a squared bracket for the first problem; use a no-intersection case for the second; reinforce ± square-root branches for the third; and draw the centre-to-midpoint right triangle for the fourth.
After repair, use a different pair of circles to test whether the student can find the line and length independently.
Practice questions with checked answers
Question one: find the common chord of x² + y² = 25 and (x − 4)² + y² = 25. Answer: x = 2; endpoints (2, ±√21); length 2√21.
Question two: find the common chord of x² + y² = 25 and x² + (y − 6)² = 25. Answer: y = 3; endpoints (±4, 3); length 8.
Question three: find the common chord line of x² + y² = 25 and (x − 4)² + y² = 16. Answer: x = 25/8; length 5√39/4.
Question four: do x² + y² = 25 and (x − 8)² + y² = 1 intersect? No; the candidate line gives y² = −21/4.
Question five: find the common point of x² + y² = 25 and (x − 8)² + y² = 9. Answer: (5, 0), an external tangent contact.
For each, classify the intersection count before calculating a chord length.
A compact examination method
First read both centres and radii, and make a rough assessment of whether real intersections are possible. Second expand the necessary squared brackets and subtract one circle equation from the other.
Third simplify the resulting linear equation. If the task asks for the common-chord line, state it. If coordinates are needed, substitute the line into one original circle and solve, keeping both square-root signs when appropriate.
Fourth verify the points in both original circle equations. For chord length, use either the endpoint distance formula or 2√(r² − d²).
Finally check tangency, disjointness and concentric exceptions. A correct line of equal powers is not necessarily a positive-length chord.
Syllabus suitability and connected reading
Subtracting two circle equations is a coordinate-geometry technique that combines simultaneous equations, chord properties and distance calculations. The radical-axis terminology and some unequal-radius extensions may be enrichment depending on the student’s examination syllabus.
For Singapore, consult SEAB’s relevant 2026 O-Level Additional Mathematics list, 2027 SEC G3 list or 2027 SEC G2 list as appropriate.
Related guides include How to Find a Circle Chord Length from Its Centre and a Line, How to Reconstruct a Circle Through Three Points, How to Find a Circle Tangent to Two Parallel Lines and the Additional Mathematics archive.
Parents may enquire about eduKate Singapore small-group tutorials near Sixth Avenue MRT. The transferable skill is knowing when subtracting two nonlinear equations removes their shared structure and leaves a simpler condition. Properly taught kids shine a bright light into the future.