Additional Mathematics | How to Find a Circle Through Two Points with Its Centre on a Given Line

Two distinct points lie on infinitely many possible circles because their centres can move along the perpendicular bisector of the segment joining them. If the question also says that the centre lies on a specified straight line, intersect that line with the perpendicular bisector. The intersection, when unique, gives the centre; the distance to either given point gives the radius.

For example, a circle passes through A(0, 0) and B(4, 0), and its centre lies on y = x. The perpendicular bisector of AB is x = 2, so the centre is (2, 2). Its squared radius is 8, and its equation is (x − 2)² + (y − 2)² = 8. This guide shows both gradient and equal-distance methods, how to recognise insufficient or contradictory centre constraints, and how to verify the completed circle.

Why two points alone do not determine a circle

A circle consists of points at a fixed distance from a centre. If a circle passes through distinct points A and B, its centre O must satisfy OA = OB.

The set of all points equidistant from A and B is the perpendicular bisector of AB. That is generally an entire straight line, containing infinitely many possible centres.

Each centre on that line produces a circle through both endpoints with a corresponding radius. Therefore two ordinary points on the circumference do not determine one unique circle.

An extra independent condition is needed, such as a centre lying on another line, a known radius, a third non-collinear point or a stated diameter. The task is to identify how that extra condition restricts the centre’s possible location.

The perpendicular bisector is the main geometric tool

For two distinct points A and B, find their midpoint M. The line through M perpendicular to AB is the locus of points equidistant from A and B.

If the centre O is on this bisector, triangles OMA and OMB are congruent right triangles, so OA = OB. Conversely, if OA = OB, O must lie on that bisector.

This is why the method applies to every circle through A and B, not merely the particular circle that an examination diagram happens to show.

The next step is to combine this centre locus with the additional given straight line. Their intersection is the candidate centre. After finding it, calculate radius from an endpoint and verify both point conditions.

Read the first example’s endpoints and centre constraint

Let A(0, 0) and B(4, 0). The segment AB is horizontal, and its midpoint is M(2, 0).

A line perpendicular to a horizontal segment is vertical. Therefore the perpendicular bisector is x = 2. Every possible circle through A and B has a centre of the form (2, k).

The problem adds the condition that the centre lies on y = x. Substitute x = 2 into that line to obtain y = 2.

Thus the only candidate centre is O(2, 2). The condition has selected one point from the infinite perpendicular-bisector locus. We now have enough information to recover the radius and complete the circle equation.

Calculate the radius and equation

With centre O(2, 2), use the distance to A(0, 0). The squared radius is OA² = (2 − 0)² + (2 − 0)² = 4 + 4 = 8.

Therefore r = √8 = 2√2. The circle equation in centre-radius form is (x − 2)² + (y − 2)² = 8.

No quadratic formula is needed. The radius calculation uses ordinary coordinate distance, and its square appears directly in the equation.

A student should not write the right side as 2√2. That is the radius, while the centre-radius equation uses r² = 8. Keep those two quantities separate when presenting the final answer.

Verify both original points independently

Substitute A(0, 0) into the completed circle: (0 − 2)² + (0 − 2)² = 4 + 4 = 8. Hence A lies on the circumference.

Substitute B(4, 0): (4 − 2)² + (0 − 2)² = 4 + 4 = 8. Hence B also lies on the circumference.

Finally check the centre-line condition. The centre (2, 2) satisfies y = x because 2 = 2.

All three conditions hold. This is the appropriate verification for a reconstruction question: every supplied point and every condition on the centre must agree with the resulting circle, not merely with one intermediate line equation.

The chord midpoint is not necessarily the circle’s centre

In the opening example, the chord AB has midpoint (2, 0), while the circle’s centre is (2, 2).

Therefore AB is a chord but not a diameter. The centre lies above the chord on its perpendicular bisector.

Its perpendicular distance from the chord line y = 0 is 2. The squared radius is 8, so half the chord length is √(8 − 4) = 2, giving full chord length 4, exactly AB.

This is a good geometric consistency check. Students sometimes take the midpoint of any two points on a circle as its centre. That shortcut is valid only when the given points are endpoints of a diameter.

An equal-distance equation avoids slope calculations

Let a proposed centre be (h, k). For A(0, 0) and B(4, 0), equal squared distances give h² + k² = (h − 4)² + k².

Expand the right side: h² + k² = h² − 8h + 16 + k². Cancel h² and k² to obtain −8h + 16 = 0, so h = 2.

The specified centre line y = x gives k = h = 2. Thus O(2, 2), matching the perpendicular-bisector method.

This alternative is useful when the segment AB has an awkward gradient or the centre constraint is already given as a linear equation. Subtracting squared distances produces a linear equation without needing negative reciprocal slopes.

Derive the general equidistance equation

Let A(x₁, y₁), B(x₂, y₂) be distinct points, and let the unknown centre be O(h, k). The equality OA² = OB² is

(h − x₁)² + (k − y₁)² = (h − x₂)² + (k − y₂)².

Expand and cancel h² and k². The result is 2h(x₂ − x₁) + 2k(y₂ − y₁) = x₂² + y₂² − x₁² − y₁².

This is the linear equation of the perpendicular bisector of AB. It is valid for horizontal, vertical and sloping segments without any division by a gradient.

Combine this equation with the line on which the centre is stated to lie. If the two linear equations are independent, their solution gives a unique centre.

A second example with sloping endpoints

Suppose a circle passes through A(1, 1) and B(5, 3), and its centre lies on x + y = 7.

The chord midpoint is M((1 + 5)/2, (1 + 3)/2) = (3, 2). The chord gradient is (3 − 1)/(5 − 1) = 2/4 = 1/2.

Therefore the perpendicular-bisector gradient is −2. Its equation through (3, 2) is y − 2 = −2(x − 3), or 2x + y = 8.

The centre must satisfy both 2x + y = 8 and x + y = 7. Subtract the equations to get x = 1, then y = 6. Hence O(1, 6) is the only possible centre.

Complete and verify the sloping example

For centre O(1, 6), calculate the squared radius from A(1, 1): OA² = (1 − 1)² + (6 − 1)² = 25.

Therefore the required circle is (x − 1)² + (y − 6)² = 25.

Check B(5, 3): (5 − 1)² + (3 − 6)² = 16 + 9 = 25, so it lies on the same circumference.

The centre also satisfies x + y = 1 + 6 = 7. The independent checks confirm both point conditions and the centre restriction. This example shows that the centre need not lie near the chord midpoint, and a vertical-looking radius does not imply that AB was a diameter.

Check the second example using squared distances

Use the general equidistance equation for A(1, 1) and B(5, 3). The coordinate differences are x₂ − x₁ = 4 and y₂ − y₁ = 2.

The right-hand side is 5² + 3² − 1² − 1² = 25 + 9 − 1 − 1 = 32.

Thus 2h(4) + 2k(2) = 32, giving 8h + 4k = 32, or 2h + k = 8.

The stated centre line is h + k = 7. Subtracting gives h = 1, k = 6. This recovers the same centre without calculating any gradients. The two methods are equivalent because both encode the locus of points equidistant from A and B.

An expanded circle equation offers a third route

Write the unknown circle as x² + y² + Dx + Ey + F = 0. Its centre is (−D/2, −E/2).

For the sloping example, the centre condition x + y = 7 becomes −D/2 − E/2 = 7, or D + E = −14.

Substitute A(1, 1): 2 + D + E + F = 0, so F = 12. Substitute B(5, 3): 34 + 5D + 3E + 12 = 0, so 5D + 3E = −46.

Solve with D + E = −14 to obtain D = −2 and E = −12. Thus the circle is x² + y² − 2x − 12y + 12 = 0, which completes to (x − 1)² + (y − 6)² = 25.

When the centre line intersects the bisector once

Two non-parallel straight lines intersect in exactly one point. Therefore, if the given centre line is not parallel to the perpendicular bisector of AB, there is exactly one possible centre.

For A(0, 0) and B(4, 0), the perpendicular bisector is x = 2. The line y = x meets it once at (2, 2).

A unique centre gives a unique radius because the distance to A is fixed. It therefore gives one circle satisfying the conditions.

This is the usual situation intended by a well-posed reconstruction problem. The uniqueness comes from the independence of the two line constraints, not from a general rule that two circumference points determine a circle on their own.

A parallel centre line can make the problem impossible

Take A(0, 0) and B(4, 0), whose perpendicular bisector is x = 2. Suppose the question instead requires the centre to lie on x = 3.

The lines x = 2 and x = 3 are distinct parallel lines, so no point can belong to both. No circle centre satisfies all the given conditions.

The correct conclusion is that there is no such circle. Do not pick the point (3, 0) and simply average distances; its distances to the two supplied points are 3 and 1, so it cannot be their common circle centre.

The conflict appears before any radius calculation. This is why it helps to solve the centre constraints first and only then calculate r².

A coincident centre line leaves infinitely many circles

Again use A(0, 0) and B(4, 0). If the specified centre line is x = 2, it is exactly the perpendicular bisector.

Every point O(2, k), for real k, satisfies both centre-locus statements. Its squared radius to A is 2² + k² = 4 + k².

Therefore an infinite family of circles satisfies the data: (x − 2)² + (y − k)² = 4 + k², with real parameter k.

A second independent condition is needed to select one member of this family. The presence of two stated lines is not automatically sufficient when both describe the same geometric locus.

The diameter case is a useful special condition

If the centre is required to lie on the chord line AB itself, the only possible centre that is also on its perpendicular bisector is the midpoint M of AB.

For A(0, 0), B(4, 0) and centre line y = 0, the centre is (2, 0). Its radius is 2 and the circle equation is (x − 2)² + y² = 4.

Here AB is a diameter, so the midpoint shortcut is justified by the extra centre-line condition. Without that condition, it would not be justified.

This links ordinary coordinate reconstruction to the theorem that the midpoint of a diameter is the circle’s centre. The word “diameter” conveys additional structural information beyond simply knowing two points on the circumference.

A vertical-chord example

Let A(0, 0) and B(0, 6), and suppose the centre lies on y = x + 1. Segment AB is vertical and has midpoint (0, 3).

Its perpendicular bisector is the horizontal line y = 3. Intersect it with y = x + 1: 3 = x + 1, so x = 2. Thus the centre is O(2, 3).

The squared radius is OA² = (2 − 0)² + (3 − 0)² = 4 + 9 = 13. The circle equation is (x − 2)² + (y − 3)² = 13.

Check B: (0 − 2)² + (6 − 3)² = 4 + 9 = 13. This example reinforces that no negative-reciprocal calculation is needed for horizontal or vertical chords.

If the radius is known instead of the centre line

Suppose a circle passes through A(0, 0) and B(4, 0) and has radius 5. Its centre still lies on x = 2, so write O(2, k).

The squared radius condition gives 4 + k² = 25. Thus k² = 21, and k = √21 or −√21.

There are two circles: (x − 2)² + (y − √21)² = 25 and (x − 2)² + (y + √21)² = 25.

The known radius has not selected which side of the chord contains the centre. An extra condition such as centre above the x-axis selects the positive k. This contrasts with a non-parallel centre-line condition that often gives one centre directly.

A radius shorter than half the chord is impossible

The distance AB is 4 in the horizontal example, so half the chord length is 2. Any circle containing the two points must have radius at least 2.

If a question claims the radius is 1, the squared-distance condition would be 4 + k² = 1, giving k² = −3. There is no real centre.

The geometric reason is simpler: a circle of radius 1 has diameter 2, too short to connect two points four units apart on its circumference.

This is an efficient consistency check. Before performing coordinate calculations, compare the required radius with half the distance between the supplied points whenever the radius is given.

The centre cannot equal a given circumference point

For a genuine circle with positive radius, its centre is not on its circumference. If O were equal to A, the distance OA would be zero, forcing the radius to be zero.

With two distinct given points A and B, equal distances OA = OB make such a zero-radius centre impossible because OB would be positive.

Thus the computed intersection of the centre conditions must produce a positive radius when A and B are distinct and the equations are consistent.

This is another reason why distinct point inputs matter. If the two given points are identical, their perpendicular bisector is not determined and the problem needs different additional information.

An extra point can be a check rather than a new calculation

After finding a circle from A, B and a centre line, a question may provide a third point C and ask whether it lies on the circumference.

For the first circle (x − 2)² + (y − 2)² = 8, consider C(4, 4). Substitution gives (4 − 2)² + (4 − 2)² = 4 + 4 = 8, so C lies on the circle.

But D(2, 2) gives zero on the left rather than 8; it is the centre, not a point on the circumference.

This follow-up is a verification problem rather than another reconstruction. Students should recognise when a function or geometric object is already fully determined and use the completed equation to test later conditions.

The shortest reliable examination routine

First record the two distinct points and calculate their midpoint. Find the perpendicular bisector either by gradient and midpoint or by equal squared distances.

Second write the given line on which the centre must lie. Solve the two linear equations to obtain the centre, taking note if they are parallel or coincide.

Third calculate the squared radius from either original point and write the centre-radius circle equation. Fourth verify the other point and the centre-line condition.

If the problem asks for uniqueness, explain that two non-parallel lines determine one centre. If the constraints are dependent or contradictory, report infinitely many circles or no circle rather than inventing an isolated answer.

Practice questions with checked answers

Question one: a circle passes through (0, 0) and (4, 0), with centre on y = x. Answer: centre (2, 2), circle (x − 2)² + (y − 2)² = 8.

Question two: the same endpoints, centre on y = 0. Answer: centre (2, 0), circle (x − 2)² + y² = 4.

Question three: the same endpoints, centre on x = 3. Answer: no such circle.

Question four: the same endpoints, centre on x = 2. Answer: infinitely many circles, (x − 2)² + (y − k)² = 4 + k².

Question five: a circle passes through (1, 1) and (5, 3), with centre on x + y = 7. Answer: (x − 1)² + (y − 6)² = 25.

Question six: a circle passes through (0, 0) and (0, 6), with centre on y = x + 1. Answer: (x − 2)² + (y − 3)² = 13.

Attempt each without the model, and check both point substitutions.

This topic combines midpoints, gradients, perpendicular bisectors, simultaneous linear equations and circle reconstruction. It can be taught at different levels of depth depending on the student’s school course; some uniqueness and family-of-circles discussions are valuable enrichment.

Singapore families can consult SEAB’s relevant 2026 O-Level Additional Mathematics list, 2027 SEC G3 list or 2027 SEC G2 list and follow the school’s topic sequence.

Parents may enquire about eduKate Singapore’s maximum three-student tutorials near Sixth Avenue MRT. The enduring lesson is to recognise what additional independent information is needed to move from a family of circles to one exact equation. Properly taught kids shine a bright light into the future.