A circle tangent to two distinct parallel straight lines has a radius equal to half the perpendicular distance between those lines. Its centre lies on the straight line halfway between them. That gives the radius and a family of possible centres; one additional independent condition is usually needed to specify a unique circle.
For example, lines y = 1 and y = 7 are six units apart. Every circle tangent to both has radius 3 and centre somewhere on y = 4. If the centre also has x-coordinate 2, the equation is (x − 2)² + (y − 4)² = 9. This guide explains horizontal, vertical and sloping lines, uses exact perpendicular-distance checks, and shows when an extra point gives one, two or no circles.
Tangency means the shortest distance equals the radius
A straight line is tangent to a circle when it meets the circumference at exactly one point. The radius drawn from the centre to that contact point is perpendicular to the tangent line.
Therefore the perpendicular distance from the circle’s centre to a tangent line equals the radius. This is a more useful starting condition than guessing where a circle might fit on a sketch.
If one circle is tangent to two lines, the centre must have the same perpendicular distance from both lines. For distinct parallel lines, the points equidistant from them form their midway parallel line.
That geometric fact gives us a centre locus and a radius before any circle equation is written. The key is to distinguish what the two tangency conditions determine and what additional information is needed to select a specific centre.
The distance between parallel lines becomes the diameter
Consider two distinct parallel lines separated by perpendicular distance d. A circle tangent to both must have its centre halfway between them, so the perpendicular distance from the centre to either line is d/2.
That distance is the radius r. Therefore r = d/2 and the diameter 2r equals d.
This relationship is exact, and it applies regardless of whether the lines are horizontal, vertical or sloping. The phrase “distance between parallel lines” always means shortest perpendicular distance, not a horizontal or vertical separation measured in an arbitrary direction.
Students should identify the perpendicular gap first. Once r is found, the circle equation needs the centre coordinates. The two lines alone usually leave the centre free to slide parallel to them, so a unique equation is not yet determined.
Worked example with horizontal lines
Find circles tangent to both y = 1 and y = 7. The lines are horizontal and six units apart vertically.
Their midway line is y = (1 + 7)/2 = 4. The radius is r = (7 − 1)/2 = 3.
Therefore every such circle has a centre (h, 4) for some real h and equation (x − h)² + (y − 4)² = 9.
There are infinitely many possible values of h. The two tangency conditions do not specify where the centre lies horizontally. If a school question supplies the centre’s x-coordinate or one contact point, that extra evidence will select a particular circle.
Writing a whole family is the correct conclusion when the problem asks for all possibilities without giving more positional information.
An extra centre coordinate produces one circle
Suppose the circle tangent to y = 1 and y = 7 has its centre on x = 2. The midway condition gives centre (2, 4).
The radius is 3, so the equation is (x − 2)² + (y − 4)² = 9. This is a unique circle under the three conditions.
The lowest point of the circle is (2, 1) and the highest point is (2, 7). Those are the points of tangency with the two horizontal lines.
Check their distance from the centre: both are three units away vertically, and the corresponding radius is perpendicular to each horizontal line. The equation and contact geometry agree exactly.
Why the two contact points form a diameter
For a circle between two horizontal tangent lines, the radii from the centre to the upper and lower contact points lie on the same vertical line, pointing in opposite directions.
Thus the segment joining the two contact points passes through the centre and has length 2r. It is a diameter.
In the example, contact points (2, 1) and (2, 7) have midpoint (2, 4) and distance 6. The midpoint is the centre and half the distance is the radius 3.
This connects the new problem to a familiar coordinate-geometry method: finding a circle from diameter endpoints. But the endpoints are established here through tangency and perpendicular radii, not simply assumed to be any two points on the circumference.
The horizontal centre position may remain undetermined
Knowing the radius does not automatically identify the centre. For y = 1 and y = 7, the circles with centres (0, 4), (2, 4) and (10, 4) are all tangent to both lines.
Their equations differ: x² + (y − 4)² = 9, (x − 2)² + (y − 4)² = 9 and (x − 10)² + (y − 4)² = 9.
All three have radius 3, but they lie at different horizontal positions. The tangent lines fix the vertical position of the centre but not its horizontal coordinate.
A question that gives only the two lines and asks for “the circle” is therefore underdetermined unless another condition is implied. In mathematical modelling, recognising insufficient information is as important as completing a calculation.
Use a contact point to identify the centre
Suppose the circle is tangent to y = 1 at the point (5, 1) and also tangent to y = 7. The radius through (5, 1) is perpendicular to y = 1 and therefore vertical.
The centre must share the contact point’s x-coordinate 5. Since the midway line is y = 4, the centre is (5, 4). The radius remains 3.
Thus the circle is (x − 5)² + (y − 4)² = 9. The upper point of contact is (5, 7).
Notice that specifying a point of tangency is stronger than specifying an arbitrary point on the circle. The contact point determines a perpendicular radius direction and hence the centre’s location.
An ordinary point on the circle can give two answers
Keep the two tangent lines y = 1 and y = 7, with radius 3 and centre (h, 4). Suppose the circle passes through P(5, 4), which is not identified as a tangent point.
Substitute into the circle equation: (5 − h)² + (4 − 4)² = 9, so (5 − h)² = 9.
There are two possibilities: 5 − h = 3 or 5 − h = −3. Hence h = 2 or h = 8.
The two circles are (x − 2)² + (y − 4)² = 9 and (x − 8)² + (y − 4)² = 9. Both pass through (5, 4) and touch the same pair of horizontal lines. A student who arbitrarily chooses one centre has lost a valid solution.
A point on the upper tangent line gives a unique circle
Suppose the circle passing between y = 1 and y = 7 is required to pass through (5, 7). Because the circle is tangent to y = 7, a point of the circle on that line must be its point of tangency.
The radius to (5, 7) is vertical, so the centre has x-coordinate 5. Its y-coordinate is 4, giving centre (5, 4) and equation (x − 5)² + (y − 4)² = 9.
The same conclusion appears by substitution into the family: (5 − h)² + (7 − 4)² = 9, which reduces to (5 − h)² = 0 and gives h = 5.
Here the two possibilities merge into one repeated possibility because the supplied point is at the extreme vertical height of every permitted circle.
Some extra points are impossible
For lines y = 1 and y = 7, every permissible circle has centre height 4 and radius 3. Its highest y-coordinate is 7 and its lowest is 1.
Suppose another condition says the circle must pass through (5, 9). This is impossible. The vertical separation between that point and any centre (h, 4) is already 5, greater than the radius 3.
Algebra confirms it: (5 − h)² + (9 − 4)² = 9 gives (5 − h)² + 25 = 9, which has no real solution.
This is a useful reasoning check. Before solving a complicated equation for a free centre coordinate, consider whether the given point lies within the vertical band that contains all possible circles.
Vertical tangent lines use the same reasoning
Consider the distinct vertical lines x = −2 and x = 4. Their horizontal perpendicular separation is 6, so the radius of a circle tangent to both is 3.
The midway vertical line is x = (−2 + 4)/2 = 1. Therefore every such circle has centre (1, k) and equation (x − 1)² + (y − k)² = 9.
If an additional condition says the centre has y-coordinate 5, the equation becomes (x − 1)² + (y − 5)² = 9. The contact points are (−2, 5) and (4, 5).
The roles of x and y are simply reversed compared with the horizontal example. The method remains perpendicular distance, midway centre locus and one independent location condition.
Sloping lines require a perpendicular distance
Suppose the tangent lines are y = x and y = x + 4. They are parallel, but their separation is not 4 units measured perpendicular to the lines.
Rewrite them as x − y = 0 and x − y = −4. The perpendicular distance between lines Ax + By = C₁ and Ax + By = C₂ is |C₂ − C₁|/√(A² + B²).
Here A = 1 and B = −1. The distance is 4/√2 = 2√2. Therefore the radius is √2, not 2.
This example teaches why a vertical intercept difference is not generally the shortest gap for sloping parallel lines. Only for horizontal lines does the intercept difference directly equal perpendicular separation.
Find the midway line for sloping tangents
For y = x and y = x + 4, their halfway parallel line is y = x + 2. Every centre of a circle tangent to both lines lies on that midline.
Choose, for example, the centre (0, 2). It belongs to y = x + 2. With radius √2, its circle equation is x² + (y − 2)² = 2.
Check the distance to y = x: rewrite it as x − y = 0. At centre (0, 2), the distance is |0 − 2|/√2 = √2.
For y = x + 4, rewrite x − y = −4. The centre distance is |−2 − (−4)|/√2 = √2. Both distances equal the radius, proving tangency.
The contact points for a sloping pair
For the circle x² + (y − 2)² = 2 between y = x and y = x + 4, the contact points are the perpendicular feet from the centre (0, 2) to each line.
On y = x, the perpendicular through the centre has slope −1 and equation y = −x + 2. Solving with y = x gives x = 1, y = 1. The first contact point is (1, 1).
On y = x + 4, solve y = −x + 2 and y = x + 4. This gives x = −1, y = 3. The second contact point is (−1, 3).
Their midpoint is (0, 2), and their distance is √[(2)² + (−2)²] = 2√2, exactly the diameter. This provides an independent coordinate check.
The general formula for any distinct parallel pair
Write two parallel lines using the same non-zero normal vector: Ax + By = C₁ and Ax + By = C₂, where (A, B) ≠ (0, 0) and C₁ ≠ C₂.
Their perpendicular distance is d = |C₂ − C₁|/√(A² + B²). A circle tangent to both has radius r = d/2 = |C₂ − C₁|/[2√(A² + B²)].
Its centre (h, k) must lie on the halfway line Ah + Bk = (C₁ + C₂)/2. One coordinate remains free unless another condition is supplied.
Once the centre is chosen, the circle equation is (x − h)² + (y − k)² = r². This framework covers horizontal, vertical and sloping cases with one geometric principle.
A numerical example with coefficients three and four
Consider 3x + 4y = 5 and 3x + 4y = 15. The common normal vector has length √(3² + 4²) = 5.
The lines are separated by |15 − 5|/5 = 2 units, so the circle radius is 1. Their midpoint locus is 3x + 4y = 10.
Choose centre (2, 1), which satisfies 3(2) + 4(1) = 10. The circle equation is (x − 2)² + (y − 1)² = 1.
The centre’s perpendicular distances to the two lines are |10 − 5|/5 = 1 and |10 − 15|/5 = 1. Both equal the radius. This is a complete verification that the proposed circle is tangent to both lines.
Be careful when line coefficients are scaled differently
The distance formula |C₂ − C₁|/√(A² + B²) assumes the two parallel equations use the same A and B coefficients. A line written 2x + 2y = 8 represents the same geometric line as x + y = 4, but their constants cannot be compared directly to a differently scaled equation.
For example, compare x + y = 1 with 2x + 2y = 8. First divide the second equation by 2 to obtain x + y = 4. Now both have normal coefficients (1, 1).
The distance is |4 − 1|/√2 = 3/√2, so the radius of a circle tangent to both is 3/(2√2).
Normalising the equations before comparing constants prevents a silent scaling error. The geometry does not change when an entire line equation is multiplied by a non-zero number.
Why a sketch can be misleading for sloping lines
On a hand-drawn coordinate plane, two lines y = x and y = x + 4 appear separated by four units in the vertical direction. But the radius of a circle tangent to both is √2, not 2.
A sketch drawn without equal axis scales can exaggerate that confusion. The right geometric measurement is the perpendicular segment between the lines, and its length must be calculated or justified.
Use a diagram to identify the midpoint strip and orientation of the contact radii, but rely on the distance formula for exact values. The circle’s radius is half the perpendicular distance.
This distinction is a transferable coordinate-geometry habit: measurement directions matter, and a difference between equation constants is not automatically an ordinary Euclidean distance.
Do not mistake a line passing through a circle for tangency
A tangent line meets a circle at exactly one point. A secant line passes through the circle and intersects its circumference at two points. Simply observing that a line passes close to or through the centre does not establish tangency.
For the circle (x − 2)² + (y − 4)² = 9, the line y = 7 is tangent because the centre is exactly three units away, equal to the radius. The line y = 4 passes through the centre and is not tangent.
The distance from the centre to a line determines the relationship: equal to radius means tangent, less than radius means secant, greater than radius means no intersection.
This provides a quick check that does not require solving a simultaneous quadratic equation for intersection coordinates every time.
What if the two lines are actually the same?
The two-line construction assumes distinct parallel lines. If the equations describe the same line, their perpendicular separation is zero, and the formula r = d/2 gives zero, not an ordinary positive-radius circle tangent to two different boundaries.
A single line can be tangent to infinitely many ordinary circles of different centres and positive radii. It does not determine a unique circle or force radius zero.
Therefore check that the parallel lines are genuinely distinct before applying the midway-strip formula. Equal constants after normalisation signal coincident lines, not a narrow band between two boundaries.
The geometric assumptions are part of the problem. A formula should never be used outside the setting in which it was derived.
A compact exam solution format
First write both lines in comparable form with the same normal coefficients. Second calculate their perpendicular separation and divide by two to obtain the radius.
Third find the equation of the halfway parallel line containing the centre. Fourth use any additional independent condition, such as a contact point, a coordinate or a point through which the circle passes, to determine the remaining centre parameter.
Fifth write the centre-radius equation and check the perpendicular distances from the proposed centre to both original lines. If the extra condition yields two centres or none, report that rather than forcing one answer.
This sequence keeps geometry and algebra aligned. It also makes clear whether the given data determine a unique circle or a family of possibilities.
Five short problems with checked answers
Question one: find the radius of any circle tangent to y = 1 and y = 7. Answer: 3, with centre on y = 4.
Question two: the same circle also has centre x-coordinate 2. Answer: (x − 2)² + (y − 4)² = 9.
Question three: a circle tangent to y = 1 and y = 7 passes through (5, 4). Give all possible centres. Answer: (2, 4) and (8, 4).
Question four: a circle tangent to x = −2 and x = 4 has centre y-coordinate 5. Answer: (x − 1)² + (y − 5)² = 9.
Question five: find a circle tangent to y = x and y = x + 4 with centre x-coordinate zero. Answer: x² + (y − 2)² = 2.
For each answer, check the two centre-to-line distances instead of relying on a visual guess.
Parents, syllabus context and further reading
This topic brings together coordinate geometry, the distance from a point to a line, perpendicular gradients, circle equations and reasoning about whether information determines one or several solutions. Some extensions may be enrichment depending on the student’s assessment level.
For official Singapore references, consult SEAB’s 2026 GCE O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232, as appropriate.
At eduKateSG, maximum three-student tutorials and generally 1.5-hour weekly lessons near Sixth Avenue MRT, subject to class suitability, allow careful diagnosis of geometry and algebra prerequisites.
Related guides include How to Find a Circle Equation from Diameter Endpoints, How to Reconstruct a Circle Through Three Points, and How to Recover Unknown Coefficients from a Tangent Line. Visit the Additional Mathematics archive or contact eduKate Singapore about tutorials. Properly taught kids shine a bright light into the future.