Three Points Fix Exactly One Circle, and Three in a Row Fix None
Learn to name every part of a circle, describe its lines of symmetry and rotational symmetry, construct the one circle through three non-collinear points, and count how many circles pass through one point or two.
How many circles pass through three given points?
Mark one dot on paper. How many circles can you draw through it? As many as you like — put the compass point anywhere and open it to reach the dot.
Mark two dots. Still infinitely many circles pass through both, though the centres are now restricted.
Mark three dots, not in a straight line. Now there is exactly one circle through all three — no freedom left at all. And if the three dots happen to lie in a straight line, there is no circle through them.
That is a striking amount of structure, and it comes from a single fact: the centre of a circle is equally distant from every point on it. Demanding three fixed distances pins the centre down completely.
A cracked dinner plate is the everyday version. Given one curved fragment, three points along its broken rim are enough to reconstruct the whole plate's size and centre — which is how a conservator works out the diameter of a pot from a shard.
This page covers the first part of the CBSE Class 9 Mathematics chapter on circles — naming the parts, the symmetry of a circle, and the counting question above with its construction.
Mark two dots. Still infinitely many circles pass through both, though the centres are now restricted.
Mark three dots, not in a straight line. Now there is exactly one circle through all three — no freedom left at all. And if the three dots happen to lie in a straight line, there is no circle through them.
That is a striking amount of structure, and it comes from a single fact: the centre of a circle is equally distant from every point on it. Demanding three fixed distances pins the centre down completely.
A cracked dinner plate is the everyday version. Given one curved fragment, three points along its broken rim are enough to reconstruct the whole plate's size and centre — which is how a conservator works out the diameter of a pot from a shard.
This page covers the first part of the CBSE Class 9 Mathematics chapter on circles — naming the parts, the symmetry of a circle, and the counting question above with its construction.
What is the difference between a sector and a segment?
A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc. One is a slice from the centre, the other is a piece cut straight across.
The parts of a circle, in order of what each needs.
- Centre — the fixed point every point of the circle is equidistant from
- Radius — a segment from the centre to the circle. All radii of a circle are equal
- Chord — a segment joining any two points on the circle
- Diameter — a chord through the centre. It equals , and it is the longest chord
- Arc — a piece of the circle itself. A chord divides the circle into a minor arc and a major arc
- Sector — the region between two radii and the arc between them. A quadrant is the sector for a right angle
- Segment — the region between a chord and one of its arcs. The minor segment goes with the minor arc
- Circumference — the whole boundary; its length is
A worked identification. In a circle of radius cm:
- the diameter is cm, and no chord can be longer
- a chord of cm is possible; a chord of cm is not, since it exceeds the diameter
- two radii at cut off a quadrant, one quarter of the region
The everyday pictures. A slice of cake cut from the centre outward is a sector. The piece left when you slice straight across the top of a roti, missing the middle, is a segment. A bicycle spoke is a radius; the rim is the circumference.
Two distinctions that questions test directly. First, an arc is part of the curve — a length — while a sector and a segment are regions with area. Naming a region as an arc is a marked error even when the shape intended is clear.
Second, a diameter is a chord but a chord need not be a diameter. Every statement proved about chords therefore applies to diameters automatically, and that inheritance gets used repeatedly in the next parts of this chapter — the right angle in a semicircle, for instance, is a chord theorem applied to the longest chord.
The parts of a circle, in order of what each needs.
- Centre — the fixed point every point of the circle is equidistant from
- Radius — a segment from the centre to the circle. All radii of a circle are equal
- Chord — a segment joining any two points on the circle
- Diameter — a chord through the centre. It equals , and it is the longest chord
- Arc — a piece of the circle itself. A chord divides the circle into a minor arc and a major arc
- Sector — the region between two radii and the arc between them. A quadrant is the sector for a right angle
- Segment — the region between a chord and one of its arcs. The minor segment goes with the minor arc
- Circumference — the whole boundary; its length is
A worked identification. In a circle of radius cm:
- the diameter is cm, and no chord can be longer
- a chord of cm is possible; a chord of cm is not, since it exceeds the diameter
- two radii at cut off a quadrant, one quarter of the region
The everyday pictures. A slice of cake cut from the centre outward is a sector. The piece left when you slice straight across the top of a roti, missing the middle, is a segment. A bicycle spoke is a radius; the rim is the circumference.
Two distinctions that questions test directly. First, an arc is part of the curve — a length — while a sector and a segment are regions with area. Naming a region as an arc is a marked error even when the shape intended is clear.
Second, a diameter is a chord but a chord need not be a diameter. Every statement proved about chords therefore applies to diameters automatically, and that inheritance gets used repeatedly in the next parts of this chapter — the right angle in a semicircle, for instance, is a chord theorem applied to the longest chord.
What symmetry does a circle have, and what does it prove?
Every diameter is a line of symmetry, so a circle has infinitely many — and it looks unchanged after a rotation about the centre through any angle at all.
Why every diameter works. Fold the circle along a diameter. Each point on one side lands on a point the same distance from the centre on the other, because all radii are equal. So the two halves match exactly. Since a diameter can be drawn in any direction, the lines of symmetry are unlimited in number.
Compare this with the polygons. A square has lines of symmetry and rotational symmetry of order — it looks the same after turns of , , and . A regular hexagon has of each. A circle is the limiting case, with no smallest angle of rotation that works: works, and so does half of that.
Using symmetry to prove a property. Take a chord and let be its midpoint. Fold along the line , where is the centre. Since and (both radii), the fold carries onto — so is a line of symmetry for the figure, and the two angles at are equal. Two equal angles on a straight line are each:
The line from the centre to the midpoint of a chord is perpendicular to it, proved from symmetry alone without any calculation. The next part of this chapter proves the same result by congruent triangles and uses it constantly.
A worked consequence. In a circle of radius cm with a chord of cm, the perpendicular from the centre bisects the chord into two halves of cm, creating a right triangle:
The chord is cm from the centre — an answer that depends entirely on the bisection being exact.
Symmetry arguments are persuasive but need care. The fold above works because and are equal, which is a property of the circle. Folding is not a proof by itself; it is a proof once you can say which lengths the fold matches and why. In a written answer, name the equal radii — that sentence is what turns the picture into an argument.
Why every diameter works. Fold the circle along a diameter. Each point on one side lands on a point the same distance from the centre on the other, because all radii are equal. So the two halves match exactly. Since a diameter can be drawn in any direction, the lines of symmetry are unlimited in number.
Compare this with the polygons. A square has lines of symmetry and rotational symmetry of order — it looks the same after turns of , , and . A regular hexagon has of each. A circle is the limiting case, with no smallest angle of rotation that works: works, and so does half of that.
Using symmetry to prove a property. Take a chord and let be its midpoint. Fold along the line , where is the centre. Since and (both radii), the fold carries onto — so is a line of symmetry for the figure, and the two angles at are equal. Two equal angles on a straight line are each:
The line from the centre to the midpoint of a chord is perpendicular to it, proved from symmetry alone without any calculation. The next part of this chapter proves the same result by congruent triangles and uses it constantly.
A worked consequence. In a circle of radius cm with a chord of cm, the perpendicular from the centre bisects the chord into two halves of cm, creating a right triangle:
The chord is cm from the centre — an answer that depends entirely on the bisection being exact.
Symmetry arguments are persuasive but need care. The fold above works because and are equal, which is a property of the circle. Folding is not a proof by itself; it is a proof once you can say which lengths the fold matches and why. In a written answer, name the equal radii — that sentence is what turns the picture into an argument.
How do you construct the circle through three given points?
Draw the perpendicular bisectors of two of the three joining segments; where they cross is the centre.
The reasoning. A point is equidistant from and exactly when it lies on the perpendicular bisector of . So the centre — which must be equidistant from all three — lies on the perpendicular bisector of and on that of . Two lines meet at one point, and that point is the centre. Its distance to any of the three is the radius.
Worked construction with coordinates. Take , and .
- runs along the -axis from to , so its perpendicular bisector is the vertical line
- runs up the -axis from to , so its perpendicular bisector is the horizontal line
- They cross at the centre
The radius is the distance from to :
Check the other two points. To : . To : . All three are exactly from , so one circle of radius passes through all three — and only one, since the centre was forced.
A bonus the check reveals. The distance is , which is . So is a diameter of this circle, and the angle at is the right angle between the axes. That is the semicircle result of the third part of this chapter, appearing here as arithmetic.
Why three collinear points have no circle. Take , , , all on a line. The perpendicular bisector of is and that of is . Two distinct vertical lines never meet, so there is no point equidistant from all three, and hence no circle.
The general reason is the same: for collinear points every perpendicular bisector is perpendicular to the same line, so they are all parallel to each other and cannot intersect.
The compass-and-ruler version. Join to and to . Bisect each segment perpendicularly with arcs from both end points, using a radius more than half the segment. Mark the intersection as , set the compass to , and draw. Use the third point as a check, not as a construction step — if the circle misses , a bisector was drawn inaccurately, and that check is the reason to plot all three.
The reasoning. A point is equidistant from and exactly when it lies on the perpendicular bisector of . So the centre — which must be equidistant from all three — lies on the perpendicular bisector of and on that of . Two lines meet at one point, and that point is the centre. Its distance to any of the three is the radius.
Worked construction with coordinates. Take , and .
- runs along the -axis from to , so its perpendicular bisector is the vertical line
- runs up the -axis from to , so its perpendicular bisector is the horizontal line
- They cross at the centre
The radius is the distance from to :
Check the other two points. To : . To : . All three are exactly from , so one circle of radius passes through all three — and only one, since the centre was forced.
A bonus the check reveals. The distance is , which is . So is a diameter of this circle, and the angle at is the right angle between the axes. That is the semicircle result of the third part of this chapter, appearing here as arithmetic.
Why three collinear points have no circle. Take , , , all on a line. The perpendicular bisector of is and that of is . Two distinct vertical lines never meet, so there is no point equidistant from all three, and hence no circle.
The general reason is the same: for collinear points every perpendicular bisector is perpendicular to the same line, so they are all parallel to each other and cannot intersect.
The compass-and-ruler version. Join to and to . Bisect each segment perpendicularly with arcs from both end points, using a radius more than half the segment. Mark the intersection as , set the compass to , and draw. Use the third point as a check, not as a construction step — if the circle misses , a bisector was drawn inaccurately, and that check is the reason to plot all three.
How many circles pass through one point, or through two?
Through one point: infinitely many. Through two points: infinitely many, but with the centres confined to one line. Through three non-collinear points: exactly one.
**Through one point .** Any point can be the centre; set the radius to . Since is unrestricted, the circles are unlimited in number and unlimited in size.
**Through two points and .** The centre must satisfy , so it lies on the **perpendicular bisector of — and any point on that line will do.
Worked example.** Take and . The perpendicular bisector is , so every centre has the form .
- gives centre and radius — the smallest possible circle, with as its diameter
- gives centre and radius
- gives centre and radius again, the mirror image
- gives radius , larger still
Every one of these passes through both and , and there are as many as there are values of .
The smallest is worth naming. As grows the radius grows, so the minimum comes at , where the radius is half of . The smallest circle through two points has those points as the ends of a diameter — a fact that follows from the diameter being the longest chord, read the other way round.
Through three non-collinear points. Adding the third condition forces the centre onto a second perpendicular bisector, and two non-parallel lines meet exactly once. So the centre is unique, the radius is determined, and there is exactly one circle. This circle is the circumcircle of the triangle formed by the three points, and its centre is the circumcentre.
Counting the freedom explains the whole pattern. A circle needs three numbers to describe it: two for the centre and one for the radius. Each point it must pass through imposes one condition. One point leaves two degrees of freedom, two points leave one, and three points leave none — which is why the answers run infinitely many, infinitely many, exactly one.
And it explains the collinear failure. Three collinear points impose three conditions that are inconsistent rather than merely restrictive: the first two force the centre onto and the last two force it onto . Having exactly enough conditions does not guarantee a solution — they must also be compatible, which is why the theorem says non-collinear and not just three.
**Through one point .** Any point can be the centre; set the radius to . Since is unrestricted, the circles are unlimited in number and unlimited in size.
**Through two points and .** The centre must satisfy , so it lies on the **perpendicular bisector of — and any point on that line will do.
Worked example.** Take and . The perpendicular bisector is , so every centre has the form .
- gives centre and radius — the smallest possible circle, with as its diameter
- gives centre and radius
- gives centre and radius again, the mirror image
- gives radius , larger still
Every one of these passes through both and , and there are as many as there are values of .
The smallest is worth naming. As grows the radius grows, so the minimum comes at , where the radius is half of . The smallest circle through two points has those points as the ends of a diameter — a fact that follows from the diameter being the longest chord, read the other way round.
Through three non-collinear points. Adding the third condition forces the centre onto a second perpendicular bisector, and two non-parallel lines meet exactly once. So the centre is unique, the radius is determined, and there is exactly one circle. This circle is the circumcircle of the triangle formed by the three points, and its centre is the circumcentre.
Counting the freedom explains the whole pattern. A circle needs three numbers to describe it: two for the centre and one for the radius. Each point it must pass through imposes one condition. One point leaves two degrees of freedom, two points leave one, and three points leave none — which is why the answers run infinitely many, infinitely many, exactly one.
And it explains the collinear failure. Three collinear points impose three conditions that are inconsistent rather than merely restrictive: the first two force the centre onto and the last two force it onto . Having exactly enough conditions does not guarantee a solution — they must also be compatible, which is why the theorem says non-collinear and not just three.
Exam tip
Exam tip: bisect two segments, then use the third point as your check
For the circumcircle, bisect only two of the three segments — the third bisector passes through the same point and drawing it wastes time. Use the third point as a check instead.
Show the arcs. Perpendicular-bisector construction marks must stay visible; an erased construction loses the marks even with a correct circle.
Set the compass wider than half the segment when bisecting, or the arcs will not cross.
Say why the method works in one line: the centre is equidistant from all three, so it lies on both perpendicular bisectors.
For collinear points, name the reason — all the perpendicular bisectors are perpendicular to the same line, so they are parallel and never meet.
Learn the three counts: through one point, infinitely many; through two, infinitely many with centres on the perpendicular bisector; through three non-collinear, exactly one.
The smallest circle through two points has them as a diameter.
Keep sector and segment apart: sector = two radii + arc; segment = chord + arc. An arc is a length, not a region.
A diameter is a chord — the longest one, equal to — so every chord result applies to it.
And when a symmetry or folding argument is used, name the equal radii; the fold is a proof only once you say what it matches.
Show the arcs. Perpendicular-bisector construction marks must stay visible; an erased construction loses the marks even with a correct circle.
Set the compass wider than half the segment when bisecting, or the arcs will not cross.
Say why the method works in one line: the centre is equidistant from all three, so it lies on both perpendicular bisectors.
For collinear points, name the reason — all the perpendicular bisectors are perpendicular to the same line, so they are parallel and never meet.
Learn the three counts: through one point, infinitely many; through two, infinitely many with centres on the perpendicular bisector; through three non-collinear, exactly one.
The smallest circle through two points has them as a diameter.
Keep sector and segment apart: sector = two radii + arc; segment = chord + arc. An arc is a length, not a region.
A diameter is a chord — the longest one, equal to — so every chord result applies to it.
And when a symmetry or folding argument is used, name the equal radii; the fold is a proof only once you say what it matches.
Did you know
Why a broken plate still remembers how big it was
A curved fragment of pottery carries enough information to reconstruct the whole vessel, and the reason is the counting from this page: three points on the rim fix the circle completely.
The method is exactly the construction. Mark three points along the fragment's curved edge, join them in pairs, bisect two of the joins perpendicularly, and the crossing point is the centre of the original rim. Measuring from there to the fragment gives the radius, and gives the circumference of a pot nobody has seen whole.
What makes it work is that the three conditions use up all the freedom a circle has. A shard tells you nothing about the pot's height or its handles, but it cannot help telling you its diameter.
The same counting has a sharper edge in the failure case. If the three chosen points are nearly collinear — which happens when the fragment is small and barely curved — the two perpendicular bisectors meet at a very shallow angle, and a tiny error in marking a point swings the crossing point a long way. The reconstruction stops being reliable, and the reason is visible in the geometry rather than in the measuring: near-parallel lines have an ill-determined intersection.
So a wide, strongly curved shard gives a confident diameter and a small flat one gives a guess. Archaeologists prefer rim fragments for this reason, and they look for the most curved piece available.
It is a good illustration of what the theorem actually promises. Exactly one circle is a statement about existence and uniqueness, not about how easy that circle is to find — and the collinear case is not an isolated exception but the extreme end of a whole range where the construction gets progressively worse.
The method is exactly the construction. Mark three points along the fragment's curved edge, join them in pairs, bisect two of the joins perpendicularly, and the crossing point is the centre of the original rim. Measuring from there to the fragment gives the radius, and gives the circumference of a pot nobody has seen whole.
What makes it work is that the three conditions use up all the freedom a circle has. A shard tells you nothing about the pot's height or its handles, but it cannot help telling you its diameter.
The same counting has a sharper edge in the failure case. If the three chosen points are nearly collinear — which happens when the fragment is small and barely curved — the two perpendicular bisectors meet at a very shallow angle, and a tiny error in marking a point swings the crossing point a long way. The reconstruction stops being reliable, and the reason is visible in the geometry rather than in the measuring: near-parallel lines have an ill-determined intersection.
So a wide, strongly curved shard gives a confident diameter and a small flat one gives a guess. Archaeologists prefer rim fragments for this reason, and they look for the most curved piece available.
It is a good illustration of what the theorem actually promises. Exactly one circle is a statement about existence and uniqueness, not about how easy that circle is to find — and the collinear case is not an isolated exception but the extreme end of a whole range where the construction gets progressively worse.
Exam relevance
How do circle basics feed into JEE Main?
Because the vocabulary is assumed without restatement in every later circle chapter, and the circumcentre construction becomes a routine coordinate-geometry calculation.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections, examined in JEE Main. The centre-and-radius description used here becomes the equation
and the statement that three non-collinear points determine a circle becomes the standard exercise of finding the equation of the circle through three given points — solved by exactly the reasoning on this page, either by two perpendicular bisectors or by substituting the three points into the general equation and solving for three unknowns. The three unknowns are the three degrees of freedom counted in the last section.
The collinear case reappears as a degenerate condition. In Class 11, three collinear points make the resulting equations inconsistent, and recognising that before grinding through the algebra saves the whole question. Assertion-reason items on when a circle exists through given points rest on it.
The perpendicular bisector is itself a reused tool. Class 11 Straight Lines treats the locus of points equidistant from two fixed points, which is this line, and Class 11 Conic Sections generalises the equidistance idea: a circle is the locus of points at a fixed distance from one point, and changing that condition slightly gives the parabola, ellipse and hyperbola. Every conic is a locus defined by distances, and the circle is where that pattern starts.
Where the symmetry goes. The infinitely many lines of symmetry make the circle the most symmetric plane figure, which is why it appears whenever a physical problem has no preferred direction — circular orbits and wavefronts in Class 11 Physics, both examined in JEE Main and NEET.
What the questions look like. For board work, expect name the parts in a given figure, construct the circle through three points with arcs shown, state how many circles pass through one or two points, and explain why collinear points admit none — often as a one-mark reason. For JEE Main, the direct form is the equation of a circle through three points, or the centre and radius from a given equation.
How board and competitive emphasis differ. A board paper rewards the construction with visible arcs and the correctly named parts. A competitive paper never asks for a construction — it asks for the centre and radius as numbers, and the geometric insight is what shortens the algebra.
The single trap that costs the most marks. Confusing sector with segment in a figure question, and calling a region an arc. The words look interchangeable and are not: a sector is cut by two radii, a segment by a chord, and an arc is a length rather than a region. Area questions in the later parts of this chapter give different formulas for the two regions, so the naming error propagates into a wrong formula.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections, examined in JEE Main. The centre-and-radius description used here becomes the equation
and the statement that three non-collinear points determine a circle becomes the standard exercise of finding the equation of the circle through three given points — solved by exactly the reasoning on this page, either by two perpendicular bisectors or by substituting the three points into the general equation and solving for three unknowns. The three unknowns are the three degrees of freedom counted in the last section.
The collinear case reappears as a degenerate condition. In Class 11, three collinear points make the resulting equations inconsistent, and recognising that before grinding through the algebra saves the whole question. Assertion-reason items on when a circle exists through given points rest on it.
The perpendicular bisector is itself a reused tool. Class 11 Straight Lines treats the locus of points equidistant from two fixed points, which is this line, and Class 11 Conic Sections generalises the equidistance idea: a circle is the locus of points at a fixed distance from one point, and changing that condition slightly gives the parabola, ellipse and hyperbola. Every conic is a locus defined by distances, and the circle is where that pattern starts.
Where the symmetry goes. The infinitely many lines of symmetry make the circle the most symmetric plane figure, which is why it appears whenever a physical problem has no preferred direction — circular orbits and wavefronts in Class 11 Physics, both examined in JEE Main and NEET.
What the questions look like. For board work, expect name the parts in a given figure, construct the circle through three points with arcs shown, state how many circles pass through one or two points, and explain why collinear points admit none — often as a one-mark reason. For JEE Main, the direct form is the equation of a circle through three points, or the centre and radius from a given equation.
How board and competitive emphasis differ. A board paper rewards the construction with visible arcs and the correctly named parts. A competitive paper never asks for a construction — it asks for the centre and radius as numbers, and the geometric insight is what shortens the algebra.
The single trap that costs the most marks. Confusing sector with segment in a figure question, and calling a region an arc. The words look interchangeable and are not: a sector is cut by two radii, a segment by a chord, and an arc is a length rather than a region. Area questions in the later parts of this chapter give different formulas for the two regions, so the naming error propagates into a wrong formula.
Key takeaways
Parts of a circle, symmetry and the circle through three points: quick revision
- The centre is equidistant from every point of the circle; all radii are equal.
- A chord joins two points on the circle; a diameter is a chord through the centre, equal to , and it is the longest chord.
- A chord splits the circle into a minor and a major arc. An arc is a length.
- A sector is bounded by two radii and an arc; a segment by a chord and an arc. Both are regions.
- A quadrant is the sector for a right angle; the circumference is .
- In a circle of radius cm the diameter is cm, so a chord of cm is impossible.
- Every diameter is a line of symmetry, so a circle has infinitely many; it is unchanged by a rotation through any angle about the centre.
- A square has lines and order ; a regular hexagon and ; a circle is the limiting case with no smallest angle.
- Symmetry proves that the line from the centre to a chord's midpoint is perpendicular to the chord — because and .
- Radius cm with a chord of cm gives distance cm from the centre.
- To construct the circle through three points, bisect two of the joining segments perpendicularly; the crossing point is the centre.
- With , , : bisectors and meet at , and — confirmed for all three points.
- In that circle , so is a diameter and the angle at is a right angle.
- Three collinear points have no circle: for , , the bisectors and are parallel and never meet.
- Through one point: infinitely many circles. Through two: infinitely many, centres on the perpendicular bisector of the join.
- For and , centres are : gives , gives , gives .
- The smallest circle through two points has them as a diameter — at , radius half of .
- Through three non-collinear points: exactly one — the circumcircle, centred at the circumcentre.
- A circle has three degrees of freedom (centre and radius), and each point imposes one condition — which is why the counts run infinitely many, infinitely many, exactly one.
Find a round object at home, mark three points on its rim through a tracing, and reconstruct its centre with two perpendicular bisectors — then measure to check you found the real middle.
- A chord joins two points on the circle; a diameter is a chord through the centre, equal to , and it is the longest chord.
- A chord splits the circle into a minor and a major arc. An arc is a length.
- A sector is bounded by two radii and an arc; a segment by a chord and an arc. Both are regions.
- A quadrant is the sector for a right angle; the circumference is .
- In a circle of radius cm the diameter is cm, so a chord of cm is impossible.
- Every diameter is a line of symmetry, so a circle has infinitely many; it is unchanged by a rotation through any angle about the centre.
- A square has lines and order ; a regular hexagon and ; a circle is the limiting case with no smallest angle.
- Symmetry proves that the line from the centre to a chord's midpoint is perpendicular to the chord — because and .
- Radius cm with a chord of cm gives distance cm from the centre.
- To construct the circle through three points, bisect two of the joining segments perpendicularly; the crossing point is the centre.
- With , , : bisectors and meet at , and — confirmed for all three points.
- In that circle , so is a diameter and the angle at is a right angle.
- Three collinear points have no circle: for , , the bisectors and are parallel and never meet.
- Through one point: infinitely many circles. Through two: infinitely many, centres on the perpendicular bisector of the join.
- For and , centres are : gives , gives , gives .
- The smallest circle through two points has them as a diameter — at , radius half of .
- Through three non-collinear points: exactly one — the circumcircle, centred at the circumcentre.
- A circle has three degrees of freedom (centre and radius), and each point imposes one condition — which is why the counts run infinitely many, infinitely many, exactly one.
Find a round object at home, mark three points on its rim through a tracing, and reconstruct its centre with two perpendicular bisectors — then measure to check you found the real middle.