Did you know that the length of a common external tangent segment between two circles can be found with one right triangle? If their centres are d units apart and their radii are r₁ and r₂, then the segment joining the two tangent contact points has length L = √[d² − (r₁ − r₂)²], provided the geometry admits that common tangent.
For example, take two circles with centres O₁(0,0), O₂(13,0) and radii 8 and 3. Their centre distance is 13, and the radius difference is 5, so L = √(13² − 5²) = 12. The two direct common tangents are 5x + 12y = 104 and 5x − 12y = 104. This guide explains why the radii are subtracted for a direct tangent, why they are added for a transverse tangent, and when the geometry gives no real common tangent.
Understand what a common tangent means
A common tangent is a straight line that touches both circles without cutting through either circumference. There can be several different common tangents depending on how far apart the circles are and whether one lies inside another.
For two separate circles, the direct or external common tangents touch the circles on corresponding sides of the line joining their centres. Their contact radii point in the same general perpendicular direction toward each tangent line.
The transverse or internal common tangents pass between the circles, with the contact radii on opposite sides of the common tangent segment. Their lengths use a different formula.
Before calculating, identify which tangent type the question describes. Subtracting radii and adding radii are not interchangeable shortcuts; they encode distinct geometric arrangements.
A radius meets its tangent at ninety degrees
Let T₁ and T₂ be the contact points on the first and second circles along a direct common tangent. Each contact radius is perpendicular to the tangent line.
Thus O₁T₁ has length r₁ and O₂T₂ has length r₂, and both are perpendicular to T₁T₂.
The common tangent segment T₁T₂ is the length we want. It is not the distance between the circle centres, nor the straight gap between the nearest circumference points along the centre line.
The perpendicular radii create a right-triangle construction. Translating one radius parallel to itself reveals why the relevant short perpendicular side is the difference |r₁ − r₂| rather than the sum.
Derive the direct tangent formula
Connect the centres O₁ and O₂, separated by length d. The two perpendicular contact radii have lengths r₁ and r₂ and point toward the same tangent line.
Translate the smaller radius parallel to itself until its endpoint lies on the longer radius’s perpendicular direction. The remaining perpendicular offset between the centres is |r₁ − r₂|.
A right triangle then has hypotenuse d, one leg |r₁ − r₂| and the other leg equal to the tangent segment length L.
By Pythagoras, L² + (r₁ − r₂)² = d². Therefore L = √[d² − (r₁ − r₂)²], taking the non-negative square root because a segment length cannot be negative.
The main 5–12–13 example
Suppose the first circle has radius 8, the second has radius 3 and the centres are 13 units apart.
The difference in radii is 8 − 3 = 5. The right triangle has hypotenuse 13 and short perpendicular leg 5.
Therefore L = √(13² − 5²) = √(169 − 25) = √144 = 12.
The result is exactly twelve units. The 5–12–13 triangle provides a clear geometric explanation and an easy independent numerical check. It also shows that the direct tangent segment is shorter than the centre separation when the radii are unequal.
Check the segment length against the centre separation
In any nondegenerate direct-tangent right triangle, the centre separation d is the hypotenuse. Therefore the tangent segment length satisfies 0 < L ≤ d.
For unequal radii, the difference |r₁ − r₂| is positive, so L < d. For equal radii, the difference is zero and L = d.
In the main example, L = 12 is less than d = 13, as it must be. A calculated length such as 14 would be impossible under the specified radii and centre separation.
These bounds make a useful quick check. An algebraic mistake under the square root can often be detected before any coordinate geometry is attempted.
The two direct common tangents have equal lengths
For circles outside one another, one direct tangent generally lies above the centre-joining line and the other below it, in a suitable coordinate orientation.
Their geometric arrangements are mirror images across the line joining the centres. The radii and centre separation are unchanged.
Consequently both tangent segments have the same length √[d² − (r₁ − r₂)²]. In the worked example, both are length 12.
A student should not report two different segment lengths merely because the tangent lines have opposite gradients. Their equations differ, but reflection preserves their distances between corresponding contact points.
Set up exact coordinate equations for the worked circles
Let the first circle be C₁: x² + y² = 64, centred at O₁(0,0). Let the second be C₂: (x − 13)² + y² = 9, centred at O₂(13,0).
Their radii are 8 and 3, respectively, and their centre separation is 13.
Because 13 > 8 + 3 = 11, the circles are externally disjoint. They admit two direct common tangents and also two transverse common tangents.
To verify the direct tangent length algebraically, we can find both tangent line equations from perpendicular centre-to-line distances and then calculate their contact coordinates.
Use a line with normal coefficients five and twelve
A direct tangent in this example may be written Ax + By = C, with the same signed side of the line for both centres.
The normal vector can be chosen proportional to (5,12), because its length is 13 and its horizontal component corresponds to the radii difference 5 across centre separation 13.
For the upper tangent, choose 5x + 12y = 104. The perpendicular distance from O₁(0,0) to this line is 104/√(25 + 144) = 104/13 = 8.
The distance from O₂(13,0) is |5(13) − 104|/13 = 39/13 = 3. These equal the two radii, proving this line touches both circles.
Find the second direct tangent line
Reflect the upper tangent across the x-axis. Its equation becomes 5x − 12y = 104.
The distance from O₁ to this line is again 104/13 = 8, and its distance from O₂ is |65 − 104|/13 = 3.
Therefore 5x + 12y = 104 and 5x − 12y = 104 are both direct common tangent lines for the two circles.
They have opposite gradients, namely −5/12 and +5/12. Both are valid because their distance conditions are exactly satisfied. A rough sketch can show the symmetry, but the distance calculations provide the proof.
Find the upper tangent contact on the first circle
The upper line is 5x + 12y = 104. Its unit normal toward the tangent is (5/13,12/13).
Starting from O₁(0,0), move radius 8 in that direction. The first contact point is T₁ = 8(5/13,12/13) = (40/13,96/13).
Check that it lies on C₁: (40/13)² + (96/13)² = (1600 + 9216)/169 = 10816/169 = 64.
It also lies on the line because 5(40/13) + 12(96/13) = (200 + 1152)/13 = 104. Both conditions are exact.
Find the upper tangent contact on the second circle
The second centre is O₂(13,0) and the radius is 3. Move three units in the same unit-normal direction (5/13,12/13).
The contact point is T₂ = (13,0) + 3(5/13,12/13) = (184/13,36/13).
Its displacement from O₂ is (15/13,36/13), whose squared length is (225 + 1296)/169 = 1521/169 = 9, matching radius squared.
Substitution into 5x + 12y gives (920 + 432)/13 = 104. Thus T₂ lies on both the second circle and the upper common tangent.
Verify the tangent segment’s length by coordinates
The coordinate difference from T₁(40/13,96/13) to T₂(184/13,36/13) is (144/13,−60/13).
Its squared length is (144² + 60²)/13² = (20736 + 3600)/169 = 24336/169 = 144.
Therefore the segment length is √144 = 12, exactly the value predicted by the right-triangle formula.
This independent coordinate check is useful because it verifies the specific tangent line as well as the general geometry. If the contact points had been paired from different tangents, their coordinate distance would not reproduce the intended segment.
Prove the contact radius is perpendicular to the line
For the upper tangent, the normal vector to the line is (5,12). Each contact radius is parallel to this vector.
The displacement along the tangent segment is T₂ − T₁ = (144/13,−60/13). Its dot product with (5,12) is [5(144) + 12(−60)]/13 = (720 − 720)/13 = 0.
Thus the segment is perpendicular to both contact radii, as a common tangent must be.
This vector verification is optional enrichment. The perpendicular-distance calculation already establishes tangency, but the dot product makes the geometry of the contact segment explicit.
The transverse tangent uses the sum of radii
For an internal or transverse common tangent, the contact radii point toward opposite sides of the tangent line along the centre-joining direction.
The corresponding perpendicular offset in the right triangle is r₁ + r₂ rather than |r₁ − r₂|.
Therefore the transverse tangent segment length is L_internal = √[d² − (r₁ + r₂)²], provided d ≥ r₁ + r₂.
In the worked example, d = 13 and r₁ + r₂ = 11, so L_internal = √(169 − 121) = √48 = 4√3. This is shorter than the direct tangent segment of length 12.
Why external and internal tangent lengths differ
The two formulas come from the same right-triangle principle but use different perpendicular legs.
For a direct tangent, the radii point in the same general direction, leaving an offset of their difference. For a transverse tangent, the radii point toward opposite sides and the offsets add.
Because r₁ + r₂ ≥ |r₁ − r₂|, the transverse common tangent segment, when it exists, is no longer than the direct segment.
Students who use the sum in a direct tangent question will underestimate the length or may incorrectly claim that no tangent exists. Label the tangent type on a sketch before choosing a formula.
Separate circles can have four common tangents
If d > r₁ + r₂, the two circles are externally disjoint. They admit two direct common tangents and two transverse common tangents.
The main example satisfies 13 > 11, so all four types of straight tangent lines exist. We calculated the two direct lines explicitly and the transverse segment length through geometry.
The number of lines and the number of segment lengths are different counts. The two direct tangent segments have the same length, and the two transverse tangent segments have another common length.
A question that asks only for “a common tangent” may need further information or a diagram to decide which of the four lines is intended.
Overlapping circles can still have direct common tangents
When |r₁ − r₂| < d < r₁ + r₂, the circles intersect at two points.
In that arrangement, two direct common tangents still exist, and the length √[d² − (r₁ − r₂)²] is real and positive.
But transverse common tangents do not exist because the circles overlap. Their length formula would require the square root of d² − (r₁ + r₂)², a negative number.
This is a useful counterexample to the idea that two circles must be completely separate before they can share a direct common tangent. The correct existence conditions depend on the kind of common tangent.
A smaller circle strictly inside the larger has no common tangent
Suppose d < |r₁ − r₂|. The smaller circle lies strictly inside the larger without touching its circumference.
No straight line can be tangent to both circumferences because every line touching the inner circle passes through the interior of the larger circle, rather than just touching its edge.
The direct tangent formula reflects this impossibility: d² − (r₁ − r₂)² is negative, so no real segment length exists.
For example, if r₁ = 8, r₂ = 3 and d = 4, the radius difference is 5. Since 4 < 5, no real common tangent segment can be constructed. Do not take a complex square root as a geometric length.
Internal tangency is a boundary case
If d = |r₁ − r₂| > 0, one circle lies inside the other and touches it at a single point.
The direct tangent formula gives L = √[d² − (r₁ − r₂)²] = 0. The two direct tangent lines merge at the shared contact point.
There is one common tangent line there, but no positive-length segment joining distinct tangent contact points.
This is a degenerate case, not a pair of ordinary direct tangent segments of zero positive length. State the geometric interpretation instead of treating the equation’s repeated solution as two different lines.
External tangency is another boundary case
If d = r₁ + r₂, the circles touch externally at one point. A line through that common point perpendicular to the centre line is a shared tangent.
The transverse-segment formula gives zero at this boundary because its two contact points coincide at the mutual tangency point.
Meanwhile the direct tangent segment length is √[(r₁ + r₂)² − (r₁ − r₂)²] = 2√(r₁r₂), which is positive when both radii are positive.
This shows why the contact type and formula must be named. External circle tangency does not automatically mean every common tangent segment is zero; the direct tangents remain distinct.
Equal radii give a simple direct-tangent length
If r₁ = r₂, their difference is zero. The direct tangent formula reduces to L = √(d²) = d for nonnegative centre separation.
Geometrically, the two contact radii have equal lengths and point in the same direction, so the segment joining their contact points is parallel and equal in length to the centre-joining segment.
For example, two equal-radius circles with centres 13 units apart have direct common tangent segments exactly 13 units long, regardless of the shared radius, provided the circles are distinct and the tangent construction is meaningful.
This special case is a useful formula check. An expression that still depends on the equal radius for direct tangent length may have used the transverse formula by mistake.
Concentric circles need separate attention
If the two circles share the same centre but have different radii, a line tangent to one cannot have the different centre-to-line distance required to be tangent to the other. There is no common tangent.
If the radii and centres are both equal, the circles coincide. They share infinitely many tangent lines, but there is no unique segment joining two distinct contact points on separate circumferences.
Therefore the ordinary formula and the statement “two direct common tangents” require a nondegenerate arrangement of distinct circles.
Check coincident or concentric cases before dividing by centre separation or claiming that there must be exactly two lines.
Find a centre distance from a known direct tangent length
The Pythagorean relationship is L² = d² − (r₁ − r₂)². Rearranging gives d² = L² + (r₁ − r₂)².
Suppose the direct tangent segment is 12 units and the two radii differ by 5 units. Then d = √(12² + 5²) = √169 = 13.
This reverses the opening calculation. It is an important transfer question because a student should be able to identify the right triangle regardless of which side is unknown.
The same relationship works for any positive direct segment with meaningful radius data. Additional information may still be needed to locate the centres in a coordinate plane.
Recover a radius difference from two measurements
Suppose the centre separation is 13 units and a direct common tangent segment is 12 units.
Then (r₁ − r₂)² = d² − L² = 169 − 144 = 25, so |r₁ − r₂| = 5.
This determines the magnitude of the radius difference, not the individual radii. The ordered pairs (8,3), (9,4) and (11,6) all have difference 5 and are compatible with that direct-tangent triangle, subject to the circles’ actual arrangement.
A problem asking for both radii needs additional independent information, such as their sum or one radius. Geometry often identifies a relationship without uniquely determining every dimension.
The angle between centreline and tangent segment
In the direct-tangent right triangle, the hypotenuse has length d and the perpendicular leg is |r₁ − r₂|.
If φ is the acute angle between the centre-joining line and the direct tangent segment, then sinφ = |r₁ − r₂|/d.
For d = 13 and radius difference 5, sinφ = 5/13. Thus φ = arcsin(5/13), approximately 22.62°.
The contact-to-contact segment has direction ratio 12 to 5 in the coordinate example, consistent with tangent slopes ±5/12. Angle reasoning provides another check that the chosen line orientations match the triangle’s dimensions.
Why a line tangent to only one circle is not enough
A proposed straight line through a point near two circles may touch the first circle but cross the second or miss it completely.
For a line to be a common tangent, its perpendicular distance from each centre must equal that circle’s radius.
In the coordinate example, both proposed lines 5x ± 12y = 104 pass both distance tests: distance 8 from O₁ and distance 3 from O₂.
Checking only one circle would leave the second tangency condition unproven. A correct common-tangent solution should verify both distances or both radius-perpendicular contact relationships.
A compact exam method
First identify the two radii and the centre separation. If centres are given in coordinates, calculate d with the ordinary distance formula.
Second determine whether the task asks for a direct external tangent or transverse internal tangent. Use the radius difference for direct tangents and the radius sum for transverse tangents.
Third check that the relevant expression under the square root is nonnegative and interpret equality or nonexistence geometrically.
Fourth calculate the segment length, maintaining exact radicals where possible. If actual tangent equations are required, impose the correct point-to-line distance for both centres and verify the contact points. This is shorter and more reliable than guessing tangents on a sketch.
Practice questions with checked answers
Question one: radii 8 and 3, centre separation 13. Direct tangent segment length: 12.
Question two: same circles. Transverse tangent segment length: 4√3.
Question three: equal radii and centre separation 13. Direct tangent segment length: 13.
Question four: radii 8 and 3, centre separation 4. Does a direct common tangent exist? No, because 4 < 5.
Question five: direct tangent length 12, radius difference 5. Centre separation: 13.
Question six: centre separation 13, direct tangent length 12. Magnitude of radius difference: 5.
Question seven: circles with radii r₁ and r₂ touch externally. Direct common tangent segment length: 2√(r₁r₂), for positive radii.
In every question, draw or imagine the right triangle and explain which two radii produce the short perpendicular leg.
Syllabus fit and related reading
Common tangent geometry connects circle equations, right triangles, perpendicular distances and coordinate gradients. The general formulas for two circles are useful extensions and may be enrichment beyond the exact requirements of a student’s school course.
For Singapore, check the relevant SEAB syllabus listings for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, as appropriate.
Related guides include How to Find Both Tangents from an External Point to a Circle, How to Find a Circle Tangent to Two Parallel Lines and How to Find the Common Chord by Subtracting Two Circle Equations.
For more resources, visit the Additional Mathematics archive or the eduKate small-group tutorial reference. Families can enquire about suitable Mathematics tutorials. The enduring skill is matching the radius geometry to the right tangent type before choosing the appropriate Pythagorean relationship. Properly taught kids shine a bright light into the future.