One Arc, Two Angles, and the Centre Always Gets Double
Learn to use the angle-at-the-centre theorem, prove that the angle in a semicircle is a right angle and that angles in the same segment are equal, test whether four points are concyclic, and find angles in a cyclic quadrilateral.
Why do two different points see the same arc at the same angle?
Draw a circle and mark a chord near the bottom. Now pick any point on the major arc above it and measure . Pick a completely different point on that same arc and measure .
The two angles are equal — always, however far apart and are.
And if you measure the angle the same chord makes at the centre, it is exactly twice as big:
That single relationship generates the whole of this page. Because the central angle is fixed by the arc, every point on the remaining arc must halve it — which is why they all agree. And because a diameter subtends at the centre, every point on the circle sees a diameter at .
A goalkeeper's view of the goal mouth is the everyday version. Standing anywhere on a particular arc through the two goalposts, the goal appears to span the same angle — move along that arc and the apparent width does not change, though moving off it does.
This page covers the third part of the CBSE Class 9 Mathematics chapter on circles — the angle theorems, the concyclic test, and cyclic quadrilaterals.
The two angles are equal — always, however far apart and are.
And if you measure the angle the same chord makes at the centre, it is exactly twice as big:
That single relationship generates the whole of this page. Because the central angle is fixed by the arc, every point on the remaining arc must halve it — which is why they all agree. And because a diameter subtends at the centre, every point on the circle sees a diameter at .
A goalkeeper's view of the goal mouth is the everyday version. Standing anywhere on a particular arc through the two goalposts, the goal appears to span the same angle — move along that arc and the apparent width does not change, though moving off it does.
This page covers the third part of the CBSE Class 9 Mathematics chapter on circles — the angle theorems, the concyclic test, and cyclic quadrilaterals.
Formula
How do you use the angle at the centre to find an unknown angle?
The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle:
with the centre and on the major arc when is the minor-arc angle.
Why it is true. Join to and extend to meet the circle. In triangle , the sides and are both radii, so the triangle is isosceles and the base angles are equal — call each . The exterior angle of that triangle at is then , since an exterior angle equals the sum of the two opposite interior angles. The same argument on triangle gives for its exterior angle with base angles . Adding the two pieces,
The isosceles triangles are the whole proof, and they exist only because all radii are equal.
Worked example 1. An arc subtends at the centre. At a point on the remaining arc it subtends
Worked example 2 — the other direction. at a point on the circle, so at the centre
Worked example 3. at the centre gives for any on the major arc.
Worked example 4 — a chord equal to the radius. Such a chord subtends at the centre, as the second part of this chapter showed, so at the circumference it subtends
The point must be on the remaining arc. If sits on the minor arc instead — the same side as the chord's short arc — then the relevant central angle is the reflex one. For a chord with a minor-arc angle of , the reflex angle is , so a point on the minor arc sees the chord at .
Check the two results against each other: , exactly as the cyclic-quadrilateral rule later in this page requires for opposite angles. The two answers are both correct for their own arcs, and a question that does not say which arc the point lies on has not been fully specified — so read the diagram before halving anything.
with the centre and on the major arc when is the minor-arc angle.
Why it is true. Join to and extend to meet the circle. In triangle , the sides and are both radii, so the triangle is isosceles and the base angles are equal — call each . The exterior angle of that triangle at is then , since an exterior angle equals the sum of the two opposite interior angles. The same argument on triangle gives for its exterior angle with base angles . Adding the two pieces,
The isosceles triangles are the whole proof, and they exist only because all radii are equal.
Worked example 1. An arc subtends at the centre. At a point on the remaining arc it subtends
Worked example 2 — the other direction. at a point on the circle, so at the centre
Worked example 3. at the centre gives for any on the major arc.
Worked example 4 — a chord equal to the radius. Such a chord subtends at the centre, as the second part of this chapter showed, so at the circumference it subtends
The point must be on the remaining arc. If sits on the minor arc instead — the same side as the chord's short arc — then the relevant central angle is the reflex one. For a chord with a minor-arc angle of , the reflex angle is , so a point on the minor arc sees the chord at .
Check the two results against each other: , exactly as the cyclic-quadrilateral rule later in this page requires for opposite angles. The two answers are both correct for their own arcs, and a question that does not say which arc the point lies on has not been fully specified — so read the diagram before halving anything.
Why is the angle in a semicircle a right angle?
**Because a diameter subtends a straight angle at the centre, and half of is .
The proof in one line.** Let be a diameter and any other point on the circle. The arc subtends at the centre, since , , are collinear. By the previous theorem,
The result needs nothing new — it is the doubling theorem applied to the longest chord.
Worked example 1. is a diameter and , with on the circle. Then , so in triangle
Worked example 2 — the converse in use. In a circle of radius cm centred at , the points and are cm apart, which is . So is a diameter, and any other point on that circle — such as — must see it at . Check with the axes: is the origin, is on the -axis and on the -axis, so is indeed a right angle.
Why angles in the same segment are equal. Take any chord and two points , on the same arc. Each halves the same central angle:
So the equality is a consequence of both angles being halves of one fixed quantity, not a separate fact to learn.
Worked example 3. Chord subtends at on the major arc. Then it subtends at every point of that arc, and at the centre.
Worked example 4 — angles in the two different segments. For that same chord, a point on the minor arc sees it at . Check: .
"Same segment" is the condition that matters. Points on opposite sides of the chord give supplementary angles, not equal ones. A figure where and straddle the chord looks much like one where they do not, and quoting the equal-angles theorem there is the standard error. The test is which side of the chord the point lies on — check it on the diagram before writing the equality down.
The proof in one line.** Let be a diameter and any other point on the circle. The arc subtends at the centre, since , , are collinear. By the previous theorem,
The result needs nothing new — it is the doubling theorem applied to the longest chord.
Worked example 1. is a diameter and , with on the circle. Then , so in triangle
Worked example 2 — the converse in use. In a circle of radius cm centred at , the points and are cm apart, which is . So is a diameter, and any other point on that circle — such as — must see it at . Check with the axes: is the origin, is on the -axis and on the -axis, so is indeed a right angle.
Why angles in the same segment are equal. Take any chord and two points , on the same arc. Each halves the same central angle:
So the equality is a consequence of both angles being halves of one fixed quantity, not a separate fact to learn.
Worked example 3. Chord subtends at on the major arc. Then it subtends at every point of that arc, and at the centre.
Worked example 4 — angles in the two different segments. For that same chord, a point on the minor arc sees it at . Check: .
"Same segment" is the condition that matters. Points on opposite sides of the chord give supplementary angles, not equal ones. A figure where and straddle the chord looks much like one where they do not, and quoting the equal-angles theorem there is the standard error. The test is which side of the chord the point lies on — check it on the diagram before writing the equality down.
How do you tell whether four points lie on one circle?
If a segment subtends equal angles at two points on the same side of it, all four points are concyclic — the converse of the equal-angles theorem.
The test. Given the segment and points , on the same side:
Worked example 1. with and on the same side of . So , , , are concyclic, and the central angle for in that circle is .
Worked example 2 — a test that fails. and , same side. The angles differ, so no circle passes through all four — lies off the arc through . Whether it is inside or outside can be decided too: a larger angle means the point is nearer the segment, so is inside the circle through , , .
Worked example 3 — two right angles. with , on the same side. Then , , , are concyclic, and since the angle at the circumference is , the segment must be a diameter of that circle. The centre is therefore the midpoint of .
Check with coordinates. Take , , so the midpoint is and half of is . Then satisfies , and its distance from is . Also gives a right angle, and its distance from is . **All four points are units from , so they are concyclic exactly as predicted.
Why the same-side condition cannot be dropped.** If and are on opposite sides of , then equal angles do not make them concyclic — for a concyclic arrangement across the chord the angles must be supplementary instead. So the test comes in two forms, and using the wrong one gives a confident wrong answer.
One thing the test does not do. It never needs the circle to be drawn. Four points are declared concyclic on the strength of an angle equality alone, with no centre found and no radius computed — and that is exactly why it is useful. Finding the circle is a separate job, done by the perpendicular-bisector construction from the first part of this chapter, and it is only worth doing when the question asks for the centre or the radius.
The test. Given the segment and points , on the same side:
Worked example 1. with and on the same side of . So , , , are concyclic, and the central angle for in that circle is .
Worked example 2 — a test that fails. and , same side. The angles differ, so no circle passes through all four — lies off the arc through . Whether it is inside or outside can be decided too: a larger angle means the point is nearer the segment, so is inside the circle through , , .
Worked example 3 — two right angles. with , on the same side. Then , , , are concyclic, and since the angle at the circumference is , the segment must be a diameter of that circle. The centre is therefore the midpoint of .
Check with coordinates. Take , , so the midpoint is and half of is . Then satisfies , and its distance from is . Also gives a right angle, and its distance from is . **All four points are units from , so they are concyclic exactly as predicted.
Why the same-side condition cannot be dropped.** If and are on opposite sides of , then equal angles do not make them concyclic — for a concyclic arrangement across the chord the angles must be supplementary instead. So the test comes in two forms, and using the wrong one gives a confident wrong answer.
One thing the test does not do. It never needs the circle to be drawn. Four points are declared concyclic on the strength of an angle equality alone, with no centre found and no radius computed — and that is exactly why it is useful. Finding the circle is a separate job, done by the perpendicular-bisector construction from the first part of this chapter, and it is only worth doing when the question asks for the centre or the radius.
How do you find the angles of a cyclic quadrilateral?
**Opposite angles add to — so one angle gives its opposite immediately.
The proof.** In a cyclic quadrilateral , the diagonal splits the circle into two arcs. The angle stands on one arc and on the other, so their central angles are the minor and reflex angles for that same chord:
The two halves add to half of a full turn regardless of — which is why the result holds for every cyclic quadrilateral.
Worked example 1. In cyclic , and . Then
Check the total: , as every quadrilateral requires.
Worked example 2 — with algebra. In cyclic , and . Opposite angles:
So and . Check: .
Worked example 3 — the exterior angle. Extend side beyond . The exterior angle there is , and since as well, the two are equal. The exterior angle of a cyclic quadrilateral equals the interior opposite angle — a corollary worth quoting directly, since it saves a step.
If , the exterior angle at is , and that is also .
Worked example 4 — testing whether a quadrilateral is cyclic. A quadrilateral has angles , , , in order. The opposite pairs are and . Neither is , so the quadrilateral is not cyclic — no circle passes through all four vertices.
Which shapes are always cyclic. A rectangle is, because its opposite angles are each and sum to ; its diagonals are diameters of the circle. A square likewise. A general parallelogram is not — its opposite angles are equal, so they sum to only when both are . So the only cyclic parallelograms are rectangles, and that is a favourite one-mark question. A general trapezium is not cyclic either, though an isosceles one is.
The proof.** In a cyclic quadrilateral , the diagonal splits the circle into two arcs. The angle stands on one arc and on the other, so their central angles are the minor and reflex angles for that same chord:
The two halves add to half of a full turn regardless of — which is why the result holds for every cyclic quadrilateral.
Worked example 1. In cyclic , and . Then
Check the total: , as every quadrilateral requires.
Worked example 2 — with algebra. In cyclic , and . Opposite angles:
So and . Check: .
Worked example 3 — the exterior angle. Extend side beyond . The exterior angle there is , and since as well, the two are equal. The exterior angle of a cyclic quadrilateral equals the interior opposite angle — a corollary worth quoting directly, since it saves a step.
If , the exterior angle at is , and that is also .
Worked example 4 — testing whether a quadrilateral is cyclic. A quadrilateral has angles , , , in order. The opposite pairs are and . Neither is , so the quadrilateral is not cyclic — no circle passes through all four vertices.
Which shapes are always cyclic. A rectangle is, because its opposite angles are each and sum to ; its diagonals are diameters of the circle. A square likewise. A general parallelogram is not — its opposite angles are equal, so they sum to only when both are . So the only cyclic parallelograms are rectangles, and that is a favourite one-mark question. A general trapezium is not cyclic either, though an isosceles one is.
Exam tip
Exam tip: check which arc the point is on before you halve
Mark the centre, the chord and the point on the diagram first. Then decide whether the central angle you need is the ordinary one or the reflex one.
Halve only for a point on the REMAINING arc. A point on the minor arc uses : a chord with at the centre gives on the major arc and on the minor one.
Angles in the same segment are equal; angles in opposite segments are supplementary. Check which side of the chord the point is on before writing either down.
For a semicircle, say why: the diameter subtends at the centre, so the angle at the circumference is .
Name the isosceles triangles in the doubling proof — the two radii are what make the base angles equal, and that is where the marks are.
**In a cyclic quadrilateral, opposite angles add to ** — and check the four angles total before finishing.
The exterior angle equals the interior opposite angle — quote it and save a step.
To test for cyclic, add both pairs of opposite angles. fails, so no circle exists.
A parallelogram is cyclic only if it is a rectangle, since equal opposite angles summing to must each be .
For concyclic points, the two angles must be on the SAME side of the segment; two right angles on the same side also make that segment a diameter.
And in every angle-chase, write the reason beside each step — angle in the same segment, angle at the centre, cyclic quadrilateral — because an unjustified correct angle earns less than a justified one.
Halve only for a point on the REMAINING arc. A point on the minor arc uses : a chord with at the centre gives on the major arc and on the minor one.
Angles in the same segment are equal; angles in opposite segments are supplementary. Check which side of the chord the point is on before writing either down.
For a semicircle, say why: the diameter subtends at the centre, so the angle at the circumference is .
Name the isosceles triangles in the doubling proof — the two radii are what make the base angles equal, and that is where the marks are.
**In a cyclic quadrilateral, opposite angles add to ** — and check the four angles total before finishing.
The exterior angle equals the interior opposite angle — quote it and save a step.
To test for cyclic, add both pairs of opposite angles. fails, so no circle exists.
A parallelogram is cyclic only if it is a rectangle, since equal opposite angles summing to must each be .
For concyclic points, the two angles must be on the SAME side of the segment; two right angles on the same side also make that segment a diameter.
And in every angle-chase, write the reason beside each step — angle in the same segment, angle at the centre, cyclic quadrilateral — because an unjustified correct angle earns less than a justified one.
Did you know
Why a footballer's shooting angle has a best spot
Stand anywhere on the pitch and the goal mouth spans some angle in your view. A wide angle means more goal to aim at; a narrow one means the posts have closed up.
The theorems on this page say exactly where that angle is constant. Every point on a particular arc through the two goalposts sees the goal at the same angle — so running along that arc changes your position without changing your view of the target at all.
And it says which way the angle moves when you leave the arc. Draw a family of circles all passing through both posts. The small circles hug the goal line and give large angles; the big ones bulge far out and give small ones. Moving onto a smaller circle widens your shooting angle — which is why players cut inside rather than shooting from near the touchline.
There is a definite best spot along any straight run. Run directly towards the goal along a line parallel to the touchline and the angle grows, reaches a maximum, and then shrinks again as you get so close that the near post starts to swing behind you. The maximum happens where your line just touches one of those circles through the posts — tangentially, with no crossing — because a crossing would mean the angle is still changing.
The same geometry decides where to stand to photograph a wide building without the edges crowding in, and where a fielder should stand to see the widest view between two stumps.
What is pleasant is that none of it requires a measurement. The statement equal angles on one arc, larger angles on smaller circles is enough to answer the question, and it comes straight from the doubling theorem: a smaller circle through the same two points has a smaller radius, so the same chord subtends a bigger central angle in it, and half of a bigger angle is bigger.
The theorems on this page say exactly where that angle is constant. Every point on a particular arc through the two goalposts sees the goal at the same angle — so running along that arc changes your position without changing your view of the target at all.
And it says which way the angle moves when you leave the arc. Draw a family of circles all passing through both posts. The small circles hug the goal line and give large angles; the big ones bulge far out and give small ones. Moving onto a smaller circle widens your shooting angle — which is why players cut inside rather than shooting from near the touchline.
There is a definite best spot along any straight run. Run directly towards the goal along a line parallel to the touchline and the angle grows, reaches a maximum, and then shrinks again as you get so close that the near post starts to swing behind you. The maximum happens where your line just touches one of those circles through the posts — tangentially, with no crossing — because a crossing would mean the angle is still changing.
The same geometry decides where to stand to photograph a wide building without the edges crowding in, and where a fielder should stand to see the widest view between two stumps.
What is pleasant is that none of it requires a measurement. The statement equal angles on one arc, larger angles on smaller circles is enough to answer the question, and it comes straight from the doubling theorem: a smaller circle through the same two points has a smaller radius, so the same chord subtends a bigger central angle in it, and half of a bigger angle is bigger.
Exam relevance
How are circle angle theorems tested in JEE Main?
Because the right angle in a semicircle is the most reused single fact in coordinate geometry, and the concyclic condition becomes an algebraic test.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections and Straight Lines, examined in JEE Main. The semicircle theorem becomes the standard result that if is a diameter then the locus of points with is that circle — which in coordinates is the diameter form of the equation of a circle:
That equation is nothing but the perpendicularity of and written out, and it is a one-line way to write down a circle from the ends of a diameter. Recognising the right angle is what tells you to use it.
The concyclic test becomes a substitution. In Class 11, four points are concyclic when one circle's equation is satisfied by all four — checked by finding the circle through three and testing the fourth. The angle test from this page is the geometric reason that works, and it often decides such a question faster than the algebra.
The angle-at-the-centre theorem reappears throughout Class 11 Trigonometric Functions, where the unit circle's central angles are the arguments of every trigonometric ratio, and in the inscribed-angle arguments used for cyclic figures in olympiad and NTSE geometry.
The cyclic quadrilateral condition is used in Class 11 to decide when four points admit a common circle, and the corollary that only a rectangle among parallelograms is cyclic appears as a quick assertion-reason item.
Where the optimisation goes. The shooting-angle question in the previous section is a genuine maximisation, treated properly in Class 12 Application of Derivatives — and the geometric solution, a circle tangent to the line of approach, is the kind of insight that turns a long calculus problem into a short one.
What the questions look like. For board work, expect find the unknown angle with reasons, prove the semicircle result, prove opposite angles of a cyclic quadrilateral are supplementary, test whether four points are concyclic, and angle chases with two or three steps — every step needing its reason named. For JEE Main, the direct forms are the diameter form of a circle, the condition for concyclic points, and problems where spotting an inscribed right angle collapses the working.
How board and competitive emphasis differ. A board paper rewards the named reason at every step of an angle chase. A competitive paper rewards seeing the right angle or the equal angles instantly, because that recognition usually replaces several lines of coordinate algebra.
The single trap that costs the most marks. Halving the wrong central angle. A point on the minor arc sees the chord at half the reflex angle, not half the ordinary one — rather than for a chord subtending . The two answers are supplementary, both look reasonable, and only the diagram distinguishes them. Marking the point's arc before computing is the whole defence.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections and Straight Lines, examined in JEE Main. The semicircle theorem becomes the standard result that if is a diameter then the locus of points with is that circle — which in coordinates is the diameter form of the equation of a circle:
That equation is nothing but the perpendicularity of and written out, and it is a one-line way to write down a circle from the ends of a diameter. Recognising the right angle is what tells you to use it.
The concyclic test becomes a substitution. In Class 11, four points are concyclic when one circle's equation is satisfied by all four — checked by finding the circle through three and testing the fourth. The angle test from this page is the geometric reason that works, and it often decides such a question faster than the algebra.
The angle-at-the-centre theorem reappears throughout Class 11 Trigonometric Functions, where the unit circle's central angles are the arguments of every trigonometric ratio, and in the inscribed-angle arguments used for cyclic figures in olympiad and NTSE geometry.
The cyclic quadrilateral condition is used in Class 11 to decide when four points admit a common circle, and the corollary that only a rectangle among parallelograms is cyclic appears as a quick assertion-reason item.
Where the optimisation goes. The shooting-angle question in the previous section is a genuine maximisation, treated properly in Class 12 Application of Derivatives — and the geometric solution, a circle tangent to the line of approach, is the kind of insight that turns a long calculus problem into a short one.
What the questions look like. For board work, expect find the unknown angle with reasons, prove the semicircle result, prove opposite angles of a cyclic quadrilateral are supplementary, test whether four points are concyclic, and angle chases with two or three steps — every step needing its reason named. For JEE Main, the direct forms are the diameter form of a circle, the condition for concyclic points, and problems where spotting an inscribed right angle collapses the working.
How board and competitive emphasis differ. A board paper rewards the named reason at every step of an angle chase. A competitive paper rewards seeing the right angle or the equal angles instantly, because that recognition usually replaces several lines of coordinate algebra.
The single trap that costs the most marks. Halving the wrong central angle. A point on the minor arc sees the chord at half the reflex angle, not half the ordinary one — rather than for a chord subtending . The two answers are supplementary, both look reasonable, and only the diagram distinguishes them. Marking the point's arc before computing is the whole defence.
Key takeaways
Circle angle theorems and cyclic quadrilaterals: quick revision
- The angle at the centre is double the angle at any point on the remaining arc: .
- The proof joins to : two isosceles triangles with base angles and give exterior angles and , summing to .
- at the centre gives at the circumference; at the circumference gives at the centre; at the centre gives .
- A chord equal to the radius subtends at the centre and ** at the circumference.
- A point on the MINOR arc uses the reflex angle**: for at the centre it sees , and .
- **The angle in a semicircle is **, because a diameter subtends at the centre.
- With a diameter and : , so .
- Angles in the same segment are equal, since both halve the same central angle. Angles in opposite segments are supplementary.
- A chord subtending on the major arc gives at the centre and on the minor arc.
- Concyclic test: if with , on the same side of , then , , , lie on one circle.
- against means not concyclic, and the larger angle marks the nearer point.
- Two right angles on the same side make a diameter, with centre at its midpoint — checked for , , , , all units from .
- **Cyclic quadrilateral: opposite angles add to **, because the two inscribed angles halve and .
- , give , , totalling .
- and give , so and the angles are and .
- The exterior angle equals the interior opposite angle — for it is , which is also .
- Angles , , , give opposite sums and , so the quadrilateral is not cyclic.
- A rectangle and a square are always cyclic; the only cyclic parallelogram is a rectangle; an isosceles trapezium is cyclic.
- Among circles through two fixed points, the smaller ones give the larger inscribed angle.
Draw a circle, mark one chord, and measure the angle it makes at three different points on the same arc — then at one point across the chord, and check the two readings add to .
- The proof joins to : two isosceles triangles with base angles and give exterior angles and , summing to .
- at the centre gives at the circumference; at the circumference gives at the centre; at the centre gives .
- A chord equal to the radius subtends at the centre and ** at the circumference.
- A point on the MINOR arc uses the reflex angle**: for at the centre it sees , and .
- **The angle in a semicircle is **, because a diameter subtends at the centre.
- With a diameter and : , so .
- Angles in the same segment are equal, since both halve the same central angle. Angles in opposite segments are supplementary.
- A chord subtending on the major arc gives at the centre and on the minor arc.
- Concyclic test: if with , on the same side of , then , , , lie on one circle.
- against means not concyclic, and the larger angle marks the nearer point.
- Two right angles on the same side make a diameter, with centre at its midpoint — checked for , , , , all units from .
- **Cyclic quadrilateral: opposite angles add to **, because the two inscribed angles halve and .
- , give , , totalling .
- and give , so and the angles are and .
- The exterior angle equals the interior opposite angle — for it is , which is also .
- Angles , , , give opposite sums and , so the quadrilateral is not cyclic.
- A rectangle and a square are always cyclic; the only cyclic parallelogram is a rectangle; an isosceles trapezium is cyclic.
- Among circles through two fixed points, the smaller ones give the larger inscribed angle.
Draw a circle, mark one chord, and measure the angle it makes at three different points on the same arc — then at one point across the chord, and check the two readings add to .