Three Points Not in a Line Decide Exactly One Circle
Show that exactly one circle passes through three non-collinear points and construct it, link equal central angles to equal arcs and equal chords, and solve riders on arcs, chords and angles.
How many circles can pass through three given points?
Through one point you can draw endlessly many circles. Through two points, still endlessly many — every circle whose centre lies on the perpendicular bisector of the segment joining them.
Through three points that are not in a straight line, there is exactly one.
That sharp jump from infinitely many to exactly one is what makes this result worth proving rather than assuming, and the proof is short.
The argument. Suppose a circle passes through , and . Its centre must be the same distance from all three, since they are all on the circle.
- Being equidistant from and puts the centre on the **perpendicular bisector of **
- Being equidistant from and puts it on the **perpendicular bisector of **
Those two lines are not parallel, because and are not parallel — that is exactly what not in a straight line guarantees. Two non-parallel lines meet at exactly one point, so there is one possible centre, and with it one possible radius.
So the circle exists and is unique, and the construction is the proof carried out: bisect two of the three joining segments and their crossing point is the centre. That circle is the circumcircle of and its centre is the circumcentre.
And the collinear case fails for a visible reason. If , and lie on one line, the two perpendicular bisectors are both perpendicular to that line, so they are parallel and never meet. No point is equidistant from all three, so no circle passes through them.
This page covers the second part of the ICSE Class 9 Mathematics chapter on the circle: the circle through three points, equal arcs and central angles, equal chords and arcs, and riders combining them.
Through three points that are not in a straight line, there is exactly one.
That sharp jump from infinitely many to exactly one is what makes this result worth proving rather than assuming, and the proof is short.
The argument. Suppose a circle passes through , and . Its centre must be the same distance from all three, since they are all on the circle.
- Being equidistant from and puts the centre on the **perpendicular bisector of **
- Being equidistant from and puts it on the **perpendicular bisector of **
Those two lines are not parallel, because and are not parallel — that is exactly what not in a straight line guarantees. Two non-parallel lines meet at exactly one point, so there is one possible centre, and with it one possible radius.
So the circle exists and is unique, and the construction is the proof carried out: bisect two of the three joining segments and their crossing point is the centre. That circle is the circumcircle of and its centre is the circumcentre.
And the collinear case fails for a visible reason. If , and lie on one line, the two perpendicular bisectors are both perpendicular to that line, so they are parallel and never meet. No point is equidistant from all three, so no circle passes through them.
This page covers the second part of the ICSE Class 9 Mathematics chapter on the circle: the circle through three points, equal arcs and central angles, equal chords and arcs, and riders combining them.
How do you construct the circle through three given points?
Join the points, construct the perpendicular bisectors of two of the joins, and take their intersection as the centre.
The steps, for three non-collinear points , and :
- Join and
- Construct the perpendicular bisector of with compasses
- Construct the perpendicular bisector of
- Mark their intersection — this is the circumcentre
- With centre and radius , draw the circle. It must pass through and as well
**Check by measuring and **: both must equal . If they differ, one of your bisectors is not accurate.
Worked example — the right-angled case, which has a shortcut. Find the circumradius of a triangle with sides cm, cm and cm.
First notice that , so by the converse of the Pythagoras theorem the triangle is right-angled, with the cm side as the hypotenuse.
For a right-angled triangle the circumcentre is the mid-point of the hypotenuse, so
Why the mid-point of the hypotenuse works is a result from the quadrilaterals chapter. Complete the rectangle on the two legs, so that the hypotenuse becomes one of its diagonals. The diagonals of a rectangle are equal and bisect each other, so their common mid-point is the same distance from all four corners — and three of those corners are the vertices of the triangle. The mid-point of the hypotenuse is therefore equidistant from all three vertices, which is precisely what a circumcentre must be.
Check the numbers: with legs cm and cm placed at the origin, the mid-point of the hypotenuse is cm from each end of it by construction, and cm from the right-angle vertex, as required.
Where the circumcentre sits tells you about the triangle.
- In an acute triangle it lies inside
- In a right-angled triangle it lies on the triangle, at the mid-point of the hypotenuse
- In an obtuse triangle it lies outside the triangle
That last case surprises students who expect a centre to be inside its figure. Nothing in the construction requires it: the perpendicular bisectors meet wherever the geometry sends them, and the circle still passes through all three points.
The steps, for three non-collinear points , and :
- Join and
- Construct the perpendicular bisector of with compasses
- Construct the perpendicular bisector of
- Mark their intersection — this is the circumcentre
- With centre and radius , draw the circle. It must pass through and as well
**Check by measuring and **: both must equal . If they differ, one of your bisectors is not accurate.
Worked example — the right-angled case, which has a shortcut. Find the circumradius of a triangle with sides cm, cm and cm.
First notice that , so by the converse of the Pythagoras theorem the triangle is right-angled, with the cm side as the hypotenuse.
For a right-angled triangle the circumcentre is the mid-point of the hypotenuse, so
Why the mid-point of the hypotenuse works is a result from the quadrilaterals chapter. Complete the rectangle on the two legs, so that the hypotenuse becomes one of its diagonals. The diagonals of a rectangle are equal and bisect each other, so their common mid-point is the same distance from all four corners — and three of those corners are the vertices of the triangle. The mid-point of the hypotenuse is therefore equidistant from all three vertices, which is precisely what a circumcentre must be.
Check the numbers: with legs cm and cm placed at the origin, the mid-point of the hypotenuse is cm from each end of it by construction, and cm from the right-angle vertex, as required.
Where the circumcentre sits tells you about the triangle.
- In an acute triangle it lies inside
- In a right-angled triangle it lies on the triangle, at the mid-point of the hypotenuse
- In an obtuse triangle it lies outside the triangle
That last case surprises students who expect a centre to be inside its figure. Nothing in the construction requires it: the perpendicular bisectors meet wherever the geometry sends them, and the circle still passes through all three points.
Why do equal angles at the centre give equal arcs and equal chords?
Because the two triangles formed by the radii are congruent by SAS, and one arc can be rotated onto the other.
The chord part, proved. Let at the centre .
In and :
- (radii)
- (given)
- (radii)
So by SAS, and by CPCTC.
The arc part. Rotate the circle about its centre through the angle that carries onto . Since , the radius lands on , so the arc lands exactly on the arc . Two figures that can be made to coincide are equal, so
The converse is also true: equal arcs subtend equal angles at the centre. Rotate one arc onto the other and the radii must follow.
So three statements move together in a circle, and proving any one of them gives the other two:
- equal angles at the centre
- equal arcs
- equal chords
Worked example 1. In a circle of radius cm, a central angle of is drawn. Find the arc length and the chord.
The arc is the fraction of the circumference:
For the chord, the two radii and the chord make a triangle with cm and . The base angles are then each, so the triangle is equilateral and the chord is cm.
Notice that the chord is shorter than the arc — cm against cm — as it must be, since the straight route between two points is shorter than the curved one.
Worked example 2 — working backwards. An arc of cm is cut off in a circle of radius cm. Find the central angle.
Worked example 3 — equal angles, different circles. Two circles of radii cm and cm each have a central angle of . Are the arcs equal?
No. The arcs are of each circumference:
The theorem applies within one circle, or within circles of equal radii — never across circles of different sizes. Equal angles give equal arcs only when the radius is the same, and that condition is stated in the theorem for exactly this reason.
The chord part, proved. Let at the centre .
In and :
- (radii)
- (given)
- (radii)
So by SAS, and by CPCTC.
The arc part. Rotate the circle about its centre through the angle that carries onto . Since , the radius lands on , so the arc lands exactly on the arc . Two figures that can be made to coincide are equal, so
The converse is also true: equal arcs subtend equal angles at the centre. Rotate one arc onto the other and the radii must follow.
So three statements move together in a circle, and proving any one of them gives the other two:
- equal angles at the centre
- equal arcs
- equal chords
Worked example 1. In a circle of radius cm, a central angle of is drawn. Find the arc length and the chord.
The arc is the fraction of the circumference:
For the chord, the two radii and the chord make a triangle with cm and . The base angles are then each, so the triangle is equilateral and the chord is cm.
Notice that the chord is shorter than the arc — cm against cm — as it must be, since the straight route between two points is shorter than the curved one.
Worked example 2 — working backwards. An arc of cm is cut off in a circle of radius cm. Find the central angle.
Worked example 3 — equal angles, different circles. Two circles of radii cm and cm each have a central angle of . Are the arcs equal?
No. The arcs are of each circumference:
The theorem applies within one circle, or within circles of equal radii — never across circles of different sizes. Equal angles give equal arcs only when the radius is the same, and that condition is stated in the theorem for exactly this reason.
How do equal chords and equal arcs prove each other?
**Equal chords cut off equal arcs, and equal arcs are cut off by equal chords — with corresponding arcs meaning minor with minor and major with major.
The proof, in one direction.** Let be equal chords of a circle with centre .
In and : and (radii), and (given). So the triangles are congruent by SSS, giving . By the previous section, equal central angles give equal arcs, so the minor arcs and are equal.
And the major arcs follow automatically. The whole circumference is fixed, so if the minor arcs are equal then what is left over must be equal too:
The converse runs backwards through the same three statements: equal arcs give equal central angles, and equal central angles give equal chords by SAS.
Worked example 1. and are equal chords in a circle with centre , and . Find and the central angle of each major arc.
Equal chords subtend equal angles at the centre, so . The reflex angles corresponding to the major arcs are each
Worked example 2 — a rider on intersecting equal chords. Two equal chords and of a circle intersect at inside the circle. Prove that and .
Draw and .
- Equal chords are equidistant from the centre, so
- The perpendicular from the centre bisects the chord, so and , and these halves are equal
Now in and :
-
- (common hypotenuse)
- (shown above)
So by RHS, giving . Subtracting these equal pieces from the equal halves and gives , and adding them to the other halves gives .
Worked numbers for that rider. Take a circle of radius cm with two equal chords of cm, so each is cm from the centre. If the intersection is cm from the centre, then
So the four segments are cm and cm.
Check: cm, the full chord, for each of them — as required.
The boundary case worth naming. A chord equal to the diameter splits the circle into two equal arcs, so minor and major stop being distinguishable. For any shorter chord the two arcs are different, which is why the theorem insists on corresponding arcs: the minor arc of one chord equals the minor arc of the other, never the major arc.
The proof, in one direction.** Let be equal chords of a circle with centre .
In and : and (radii), and (given). So the triangles are congruent by SSS, giving . By the previous section, equal central angles give equal arcs, so the minor arcs and are equal.
And the major arcs follow automatically. The whole circumference is fixed, so if the minor arcs are equal then what is left over must be equal too:
The converse runs backwards through the same three statements: equal arcs give equal central angles, and equal central angles give equal chords by SAS.
Worked example 1. and are equal chords in a circle with centre , and . Find and the central angle of each major arc.
Equal chords subtend equal angles at the centre, so . The reflex angles corresponding to the major arcs are each
Worked example 2 — a rider on intersecting equal chords. Two equal chords and of a circle intersect at inside the circle. Prove that and .
Draw and .
- Equal chords are equidistant from the centre, so
- The perpendicular from the centre bisects the chord, so and , and these halves are equal
Now in and :
-
- (common hypotenuse)
- (shown above)
So by RHS, giving . Subtracting these equal pieces from the equal halves and gives , and adding them to the other halves gives .
Worked numbers for that rider. Take a circle of radius cm with two equal chords of cm, so each is cm from the centre. If the intersection is cm from the centre, then
So the four segments are cm and cm.
Check: cm, the full chord, for each of them — as required.
The boundary case worth naming. A chord equal to the diameter splits the circle into two equal arcs, so minor and major stop being distinguishable. For any shorter chord the two arcs are different, which is why the theorem insists on corresponding arcs: the minor arc of one chord equals the minor arc of the other, never the major arc.
Exam tip
How should a circle rider be written out for full marks?
**Join the radii first — almost every circle proof starts by drawing , or a perpendicular from the centre. The figure in the question is usually missing the lines that make the proof possible.
- Say radii of the same circle as the reason for every pair of equal radii. It is the most used justification in the chapter and it must be written, not assumed
- Draw the perpendicular from the centre to a chord** whenever a chord length appears, and quote the perpendicular from the centre bisects the chord
- Chain the three equivalent statements explicitly: equal chords, equal central angles, equal arcs. Write which one you are given and which one you need
- **For arcs, say minor arc or major arc.** An unqualified *arc * is ambiguous and can cost the mark
- **Use of the circumference** for an arc length, and keep when the radius is a multiple of — the arithmetic then stays exact
- For a circumcircle, construct two perpendicular bisectors and check the third distance. Three equal distances is the definition of the centre
- State the collinear exception when a question asks how many circles pass through three points: if they are in a line, none
The trap that catches strong students. Applying equal angles give equal arcs across two circles of different radii. The theorem holds in one circle or in circles of the same radius, and a question with two different circles is usually testing exactly that. Check the radii before you use it — as worked example 3 above showed, the arcs can differ by a factor of two.
- Say radii of the same circle as the reason for every pair of equal radii. It is the most used justification in the chapter and it must be written, not assumed
- Draw the perpendicular from the centre to a chord** whenever a chord length appears, and quote the perpendicular from the centre bisects the chord
- Chain the three equivalent statements explicitly: equal chords, equal central angles, equal arcs. Write which one you are given and which one you need
- **For arcs, say minor arc or major arc.** An unqualified *arc * is ambiguous and can cost the mark
- **Use of the circumference** for an arc length, and keep when the radius is a multiple of — the arithmetic then stays exact
- For a circumcircle, construct two perpendicular bisectors and check the third distance. Three equal distances is the definition of the centre
- State the collinear exception when a question asks how many circles pass through three points: if they are in a line, none
The trap that catches strong students. Applying equal angles give equal arcs across two circles of different radii. The theorem holds in one circle or in circles of the same radius, and a question with two different circles is usually testing exactly that. Check the radii before you use it — as worked example 3 above showed, the arcs can differ by a factor of two.
Did you know
How can a wheelwright rebuild a rim from three marks on the arc?
A cartwheel rim has broken and only a fragment survives. How much of a circle do you need in order to know the whole of it?
Three points. Mark any three positions along the surviving arc, construct the perpendicular bisectors of two of the chords joining them, and their meeting point is the centre of the original wheel. Measure from there to any mark and you have the radius. A fragment of arc contains the entire circle, because three points fix it and no fewer will do.
The same reasoning works on the ground. To find the centre of a circular field from three boundary stones, or to recover the curve of an old arch from three surviving bricks, you do not need the centre to be reachable or even visible — only the two bisectors.
And the right-angled case gives a carpenter a test with no measuring at all. Because the circumcentre of a right-angled triangle is the mid-point of its hypotenuse, the hypotenuse of such a triangle is a diameter of the circle through its three vertices. Turn that around: slide a set square so that its right-angle corner touches a semicircular arch while both arms pass through the two ends of the span. If the corner rides along the arc all the way across, the arch is a true semicircle; if it lifts off or digs in, it is not.
Check the idea on numbers. For legs of cm and cm the hypotenuse is cm, the circumradius is cm, and the right-angle vertex sits cm from the mid-point of the hypotenuse — on the circle, exactly as the test requires.
That single fact — a right angle standing on a diameter — is the gateway to the whole of the Class 10 circle chapter, where it becomes the angle-in-a-semicircle theorem and then the cyclic quadrilateral. You have met it here as a consequence of a rectangle's diagonals.
Three points. Mark any three positions along the surviving arc, construct the perpendicular bisectors of two of the chords joining them, and their meeting point is the centre of the original wheel. Measure from there to any mark and you have the radius. A fragment of arc contains the entire circle, because three points fix it and no fewer will do.
The same reasoning works on the ground. To find the centre of a circular field from three boundary stones, or to recover the curve of an old arch from three surviving bricks, you do not need the centre to be reachable or even visible — only the two bisectors.
And the right-angled case gives a carpenter a test with no measuring at all. Because the circumcentre of a right-angled triangle is the mid-point of its hypotenuse, the hypotenuse of such a triangle is a diameter of the circle through its three vertices. Turn that around: slide a set square so that its right-angle corner touches a semicircular arch while both arms pass through the two ends of the span. If the corner rides along the arc all the way across, the arch is a true semicircle; if it lifts off or digs in, it is not.
Check the idea on numbers. For legs of cm and cm the hypotenuse is cm, the circumradius is cm, and the right-angle vertex sits cm from the mid-point of the hypotenuse — on the circle, exactly as the test requires.
That single fact — a right angle standing on a diameter — is the gateway to the whole of the Class 10 circle chapter, where it becomes the angle-in-a-semicircle theorem and then the cyclic quadrilateral. You have met it here as a consequence of a rectangle's diagonals.
Exam relevance
How does the circle through three points appear in JEE questions?
This is foundation work whose Class 11 version is one of the most mechanical and most frequently set questions in coordinate geometry.
Where it leads. The general equation of a circle, , has three unknowns. Three points give three linear equations, and solving them gives exactly one circle — the algebraic statement of the theorem in this chapter. JEE Main asks for the equation of the circle through three given points, and the answer is unique for the same reason your construction was.
And the collinear case appears as an algebraic failure. If the three points lie on a line, the three equations become inconsistent and the determinant that would give the centre vanishes. Students who know the geometry recognise the degenerate case instantly, while those who only know the method grind through the algebra and get a contradiction.
Where the arc results lead. Equal arcs and equal central angles become the radian measure of Class 11 trigonometry, where the arc length is exactly your with the angle measured differently. From there they run into Circular Motion in JEE and NEET Physics, where angular displacement and arc length are related by the same equation.
Where the right-angle case leads. The hypotenuse of a right-angled triangle is a diameter of its circumcircle is used constantly in Class 10 circle theorems and then in coordinate geometry to write down a circle from the two ends of a diameter. The circumradius formula for a right triangle, , is worth memorising now.
Question types to expect. At this level: prove the uniqueness, construct the circle, and equal-chord or equal-arc riders. In competitive papers: the equation through three points, the circumcentre as the intersection of perpendicular bisectors, arc length in radians, and assertion-reason items on whether three given points determine a circle.
The single trap that costs marks. Forgetting that the theorem needs the points to be non-collinear, and using equal-arc results across circles of different radii. Both are conditions rather than calculations, and conditions are what assertion-reason questions are built from.
Board versus competitive emphasis. ICSE marks the construction and the equidistance argument; a competitive paper marks an equation or a length. **The idea that survives is the centre is the point equidistant from all of them** — in every later chapter, that sentence is what you actually solve.
Where it leads. The general equation of a circle, , has three unknowns. Three points give three linear equations, and solving them gives exactly one circle — the algebraic statement of the theorem in this chapter. JEE Main asks for the equation of the circle through three given points, and the answer is unique for the same reason your construction was.
And the collinear case appears as an algebraic failure. If the three points lie on a line, the three equations become inconsistent and the determinant that would give the centre vanishes. Students who know the geometry recognise the degenerate case instantly, while those who only know the method grind through the algebra and get a contradiction.
Where the arc results lead. Equal arcs and equal central angles become the radian measure of Class 11 trigonometry, where the arc length is exactly your with the angle measured differently. From there they run into Circular Motion in JEE and NEET Physics, where angular displacement and arc length are related by the same equation.
Where the right-angle case leads. The hypotenuse of a right-angled triangle is a diameter of its circumcircle is used constantly in Class 10 circle theorems and then in coordinate geometry to write down a circle from the two ends of a diameter. The circumradius formula for a right triangle, , is worth memorising now.
Question types to expect. At this level: prove the uniqueness, construct the circle, and equal-chord or equal-arc riders. In competitive papers: the equation through three points, the circumcentre as the intersection of perpendicular bisectors, arc length in radians, and assertion-reason items on whether three given points determine a circle.
The single trap that costs marks. Forgetting that the theorem needs the points to be non-collinear, and using equal-arc results across circles of different radii. Both are conditions rather than calculations, and conditions are what assertion-reason questions are built from.
Board versus competitive emphasis. ICSE marks the construction and the equidistance argument; a competitive paper marks an equation or a length. **The idea that survives is the centre is the point equidistant from all of them** — in every later chapter, that sentence is what you actually solve.
Key takeaways
What should you know about circles before moving on?
Part 2 turns the chord work of Part 1 into statements about arcs, and adds one existence theorem.
- Exactly one circle passes through three non-collinear points, because the centre must lie on two non-parallel perpendicular bisectors
- Collinear points give no circle at all — the bisectors are parallel
- Construct a circumcircle by bisecting two of the joining segments, then check the third distance
- In a right-angled triangle the circumcentre is the mid-point of the hypotenuse, so — a consequence of a rectangle's equal bisecting diagonals
- The circumcentre is inside an acute triangle, on a right-angled one, and outside an obtuse one
- Equal central angles, equal arcs and equal chords are three forms of one fact, and any one gives the other two — proved by SAS and SSS
- All of that holds within one circle or between circles of equal radii, never across different radii
- Arc length is of the circumference, and the chord is always shorter than its arc
- **Say minor or major** whenever you name an arc
The test of this chapter is the right-angled shortcut. Take a triangle with sides , and cm, find its circumradius in one line, and then explain to yourself why the mid-point of the hypotenuse is the centre — using a rectangle.
- Exactly one circle passes through three non-collinear points, because the centre must lie on two non-parallel perpendicular bisectors
- Collinear points give no circle at all — the bisectors are parallel
- Construct a circumcircle by bisecting two of the joining segments, then check the third distance
- In a right-angled triangle the circumcentre is the mid-point of the hypotenuse, so — a consequence of a rectangle's equal bisecting diagonals
- The circumcentre is inside an acute triangle, on a right-angled one, and outside an obtuse one
- Equal central angles, equal arcs and equal chords are three forms of one fact, and any one gives the other two — proved by SAS and SSS
- All of that holds within one circle or between circles of equal radii, never across different radii
- Arc length is of the circumference, and the chord is always shorter than its arc
- **Say minor or major** whenever you name an arc
The test of this chapter is the right-angled shortcut. Take a triangle with sides , and cm, find its circumradius in one line, and then explain to yourself why the mid-point of the hypotenuse is the centre — using a rectangle.