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I'm Up and Down, and Round and Round — Class 9 Mathematics

Ganita Manjari · Chapter 5

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Chapter summary

This chapter studies the circle and its many elegant properties. A circle is the set (locus) of all points in a plane at a fixed distance, the radius, from a centre. A chord joins two points on the circle; a chord through the centre is a diameter, the longest chord. Circles have rotational symmetry about the centre and reflection symmetry across every diameter. Infinitely many circles pass through two given points, with centres on the perpendicular bisector of the segment; exactly one circle (the circumcircle) passes through three non-collinear points. Key theorems: equal chords subtend equal angles at the centre and conversely; the line from the centre to a chord's midpoint is perpendicular to the chord and conversely; equal chords are equidistant from the centre and conversely; the longer of two chords lies nearer the centre. The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining circle, so the angle in a semicircle is . Four points are concyclic when a segment subtends equal angles at two points on the same side, and opposite angles of a cyclic quadrilateral sum to .

Circle and chordsSymmetries of a circleCircumcircleChord theoremsAngles and concyclicity

Key ideas & terms

Chord: A line segment joining any two points on a circle.
Diameter: A chord that passes through the centre; it is the longest chord, equal to twice the radius.
Arc: A connected portion of a circle between two points; the smaller is the minor arc, the larger the major arc.
Circumcircle: The unique circle passing through the three vertices of a triangle; its centre is the circumcentre.
Concyclic points: Points that all lie on a single circle.
Cyclic quadrilateral: A four-sided figure whose vertices all lie on one circle.

Extra questions & answers

How is a circle defined?

It is the set of all points in a plane at a fixed distance (the radius) from a given point called the centre.

How many circles pass through two given points?

Infinitely many; their centres all lie on the perpendicular bisector of the segment joining the two points.

How many circles pass through three non-collinear points?

Exactly one — the circumcircle whose centre is where the perpendicular bisectors meet.

What does the line from the centre to the midpoint of a chord do?

It is perpendicular to the chord; conversely, a perpendicular from the centre bisects the chord.

Equal chords of a circle are how placed relative to the centre?

They are equidistant from the centre; conversely, chords equidistant from the centre are equal in length.

Of two unequal chords, which is nearer the centre?

The longer chord is nearer the centre; the diameter, containing the centre, is nearest of all.

How is the angle at the centre related to the angle on the circle for the same arc?

The angle at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

What is the angle in a semicircle?

It is , since the diameter subtends a straight angle at the centre.

What is the sum of opposite angles of a cyclic quadrilateral?

They add up to .

Find the chord length if the radius is cm and the chord is cm from the centre.

Half-chord , so the full chord is cm.

Think and Reflect

Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

Amina told her to fold the circular paper so that the circular edge exactly overlaps itself. The crease is a diameter. Unfold and fold again in a different direction so that the edge again overlaps. The second crease is another diameter. The point where the two diameters intersect is the centre. Any diameter is a line of reflection symmetry, so two different diameters meet at the centre.

Think and Reflect

1. What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?

A square looks the same after rotation by , , and about its centre, so rotational symmetry of order . It has lines of reflection symmetry: the two lines joining mid-points of opposite sides and the two diagonals. A regular pentagon has rotational symmetry of , , , , (order ) and reflection lines each through a vertex and mid-point of opposite side. A regular hexagon has rotational symmetry of multiples (order ) and reflection lines: through opposite vertices and through mid-points of opposite sides.

2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?

The longest chord is the diameter through the centre. Its length is units. There is no positive smallest chord. A chord can be made arbitrarily small by taking two points very close on the circle, its length tends to . If a single point is counted as a chord of length , that is the minimum; otherwise there is no minimum positive length.

3. The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points? (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)

The locus is the perpendicular bisector of the segment . Every point on the perpendicular bisector of is equidistant from and , because triangles formed with the midpoint of are congruent by SAS: , common, , so . Conversely any point equidistant from and lies on that bisector. Hence the bisector satisfies both conditions for a locus.

Think and Reflect

1. How many circles pass through two points on a plane?

Infinitely many. The centre of any circle through and must be equidistant from and , so it must lie on the perpendicular bisector of . Every point on that bisector gives a different circle through and .

2. Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?

No, not all radii are possible. If , the radius must satisfy . The smallest circle has as diameter, so with centre at the midpoint of . There is no largest circle; by moving the centre arbitrarily far along the perpendicular bisector, can be made arbitrarily large, tending to infinity, when the circle appears almost as a straight line through and .

3. As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?

The radii increase. If is on the perpendicular bisector at distance from the midpoint of , then by Pythagoras , which grows as grows.

4. As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?

Less curved. Curvature is inversely proportional to radius . As increases, the circle becomes larger and locally flatter near and , so the arc looks less curved.

5. You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?

Infinitely many squares can have and on their boundary, because you can rotate and scale a square while keeping and on different sides. If and must be vertices (corners), there are exactly possibilities: can be a side giving two squares on opposite sides of , or can be a diagonal giving one square where is the diagonal. If order matters, that is distinct squares.

Exercise Set 5.1

1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Construct , at draw and at draw to meet at so . Draw perpendicular bisectors of and ; their intersection is the circumcentre. Join . The circle with centre and radius is the circumcircle. Since all angles are , the triangle is acute-angled, so the circumcentre lies inside the triangle.

2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Draw , at construct and mark to get . Draw perpendicular bisectors of and to locate . The circle centred at with radius is the circumcircle. Because the triangle is obtuse, so the circumcentre lies outside the triangle, on the side opposite the obtuse angle.

3. Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

Construct the triangle and its circumcircle. The distances are all equal, because the circumcentre is equidistant from the three vertices — that common length is the circumradius . For this triangle (, ), the area is , so . The triangle is acute and isosceles, so its circumcentre lies inside it.

4. What is the least possible radius of a circle through two points A and B?

The least radius is half the distance . That is , attained when is the diameter and the centre is the midpoint of . Any other circle through and has larger radius.

Think, Draw and Infer

1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?

No such point exists unless . If then lies on the perpendicular bisector of ; if then lies on the perpendicular bisector of . When are collinear, both bisectors are perpendicular to the same line , so they are parallel to each other and never meet. Hence no common intersection exists, so no point is equidistant from three collinear points. Therefore no circle can pass through three distinct collinear points, because a centre would have to be equidistant from them. Consequently a straight line can meet a circle in at most two points, never three.

2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?

Yes, infinitely many. Fix the circumcircle. Any point on the circle such that arc yields a triangle congruent to can be obtained by rotating the triangle about the centre. In particular, keep base fixed, there is a mirror point on the opposite arc where . In general, all triangles inscribed in the same circle with same side lengths are congruent by SSS.

Exercise Set 5.2

1. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Let be a chord and the centre. Then and are radii, so . Therefore has two equal sides and is isosceles with base .

2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Let triangles be and with and base . Then by SSS, , , , so . Hence equal chords give congruent central triangles.

Exercise Set 5.3

1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)

In and , (hypotenuse), is common, and . By RHS congruence, . Hence corresponding sides , so the perpendicular from the centre bisects the chord.

2. An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.

Let be midpoint of . In with , the altitude from to is also the median and perpendicular bisector of . Since the perpendicular bisector of chord passes through the centre of the circle, the line passes through . Formally, lies on the perpendicular bisector of , and is that bisector, so are collinear.

3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.

Let distances from centre to chords be . For chord cm, half-length cm, so cm. For chord cm, half-length cm, so cm. Since chords are on opposite sides, distance between midpoints is cm.

Exercise Set 5.4

1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

Let chords with midpoints and centre . Then and and , , so . By Baudhāyana-Pythagoras in right triangles and : and . Since and , we get , so , i.e., equal chords are equidistant from centre.

2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Given , and with centre . In right triangles and , , , . By RHS, , so . Since the perpendicular from centre bisects chord, and , hence .

3. Solve the previous question using the Baudhāyana–Pythagoras theorem.

From right triangles, and . Since and , we get so . Doubling gives .

Exercise Set 5.5

1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Half-length cm. So chord length .

2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is .

Let chord have midpoint , centre , so , , . In right triangle , by Baudhāyana-Pythagoras. So . Since bisects , .

3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.

No. Chord length is , which is not linear in . If , then and . In general . For example, , gives , is impossible as . Even when feasible, the ratio is not .

Exercise

A circle with centre O is drawn, and A, B, C, D are points on the circle (see Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

By measurement on the figure, the central angle subtended by arc is acute and less than (about to in the printed diagram), so is a minor arc. The central angle subtended by arc is reflex, greater than (about to ), so is a major arc. In general, measure with a protractor: minor if , major if .

Exercise Set 5.6

1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?

In , and . So is isosceles with vertex , hence equilateral. Therefore .

2. Let A and B be two points on a circle with centre O. (i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?

No. If and are on the same side of chord and both on the circle outside the arc , then by Theorem 9 the angle subtended by arc at the centre is twice the angle subtended on the remaining arc. Hence . All angles in the same segment are equal, so they cannot be different.

2. Let A and B be two points on a circle with centre O. (ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?

Not necessarily on the same side of the circle, but on the same side of line . If and both subtend segment on the same side, then lie on the same segment. However equal angles can also occur on opposite sides if the sum is for cyclic quadrilaterals. The statement is true if we specify same side of : equal subtended angles imply concyclicity on same segment.

2. Let A and B be two points on a circle with centre O. (iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?

Yes. If and are on the same side of , then by Theorem 10 the four points are concyclic. So the unique circle through also passes through . This is the converse of equal angles in same segment.

3. Find x in Fig. 5.26.

In Fig. 5.26 the angle is an inscribed angle of the cyclic quadrilateral (its vertex lies on the circle), and is the angle at the opposite vertex. Opposite angles of a cyclic quadrilateral are supplementary, so .

Exercise

A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.

Check whether the pairs of opposite angles are supplementary: and . Since both pairs of opposite angles add up to , the quadrilateral satisfies the cyclic-quadrilateral condition, so yes — such a cyclic quadrilateral can be drawn with all four vertices on one circle.

End-of-Chapter Exercises

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

Half-chord cm. So chord cm.

2. An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

By Theorem 9, angle at circumference is half the central angle, provided the point is on the remaining arc opposite the minor arc. So . If the point is on the minor arc itself, the reflex central angle is and the angle would be . Usually the minor arc gives .

3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.

Radius cm, half-chord cm. Distance cm.

4. A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?

Half-chord cm. So full chord cm.

5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Let be chord with midpoint . Suppose line is its perpendicular bisector, so at . The centre is equidistant from and (), so lies on the locus of points equidistant from and , which is the perpendicular bisector of . Hence lies on . Alternatively, the line from centre to midpoint is perpendicular to chord, so it coincides with the perpendicular bisector.

6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.

. is a diameter, so arc subtends a straight angle at centre. By Theorem 9, angle subtended by same arc at the circumference is half the central angle, so . This is the angle in a semicircle.

7. ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?

Opposite angles of a cyclic quadrilateral sum to . So . Similarly .

8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.

Since is cyclic, . So gives , , . Then , . Check .

9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.

Half-chord cm, distance cm. So cm.

10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

The quadrilateral with sides in order is a kite that is cyclic. For a cyclic quadrilateral to have two pairs of equal adjacent sides, it must be symmetric and its opposite angles each, making it consist of two right triangles with hypotenuse as diameter. The longest side pair suggests diameter. Using Brahmagupta or splitting: area is sum of two congruent right triangles with legs and . Each triangle area . Total area square units. Indeed a cyclic kite with sides has diameter and is orthogonal.

11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?

Examine the quadrilateral's angles. If all interior angles are or polygon is convex and contains the intersection of perpendicular bisectors, or equivalently if the quadrilateral can be split by a diagonal into two triangles both acute, the centre is inside. Practically, draw perpendicular bisectors of two adjacent sides; their intersection is the circumcentre. If that intersection falls inside the quadrilateral, centre is inside; if outside, it is outside. For convex cyclic quadrilaterals, the centre is inside iff the quadrilateral is not too obtuse; an equivalent test: the quadrilateral contains its circumcentre iff each side subtends an angle at opposite vertex. Obtuse angle pushes centre outside.

12. When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

Let chords and intersect at inside circle and . Equal chords are equidistant from centre, so their distances from centre are equal, where are midpoints. Triangles and are congruent right triangles, so and . Since , half-lengths . Then and give , and similarly . If intersection is at different ordering, the corresponding segments are equal.

13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)

Required radius cm, because half-chord cm and distance cm give right triangle. Construct an isosceles triangle with base cm and apex distance such that perpendicular from centre to base is cm. Simpler: draw line, mark midpoint , draw perpendicular cm to centre , then with as centre and cm draw circle; chord at distance cm will be cm. It is the circumcircle of an isosceles right triangle with legs .

14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

Let parallelogram be cyclic. Then opposite angles sum to : but in a parallelogram , so gives . Similarly all angles are . Hence it is a rectangle. Conversely every rectangle is cyclic because its diagonals are equal and bisect each other, so the intersection is equidistant from all vertices. No non-rectangular parallelogram can be cyclic.

15. Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

In a rectangle inscribed, and are diameters because angle in a semicircle is : subtended by , so passes through centre. Similarly passes through centre. The two diagonals intersect at a single point, which is the midpoint of both diameters, hence it coincides with the circle's centre.

16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

All chords of fixed length are at a fixed distance from the centre. Their midpoints are at distance from centre for every direction. Hence the locus of midpoints is a concentric circle with centre same as original and radius . If equals diameter, locus is a single point (centre); otherwise it is a circle inside.

17. In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.

If , then by Theorem 2 equal chords subtend equal angles at centre: . In triangles and , , , common, so by SSS. Hence , so bisects . Since is the line from through centre , the centre lies on the bisector.

18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Let distances from centre to chords be with for shorter chord. Then and (same side). So . Substituting : gives => => cm. Then cm.

19. A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

Central angle for each side is . Triangle formed by two radii and a side has two sides and included angle , so it is equilateral. Hence each side . Distance from centre to side (apothem) . So apothem .

20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.

Both and stand on the same chord with their vertices and on the same arc, so by the angles-in-the-same-segment theorem they are equal: . Also, since is a diameter, it subtends right angles at the other points on the circle, so .

21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).

Extend to . Then and form a linear pair, so . In a cyclic quadrilateral opposite angles are supplementary, so . Comparing the two equations, . Hence the exterior angle at a vertex equals the interior angle at the opposite vertex.

22. “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?

For any chord not through centre , triangle has sides , , opposite. By triangle inequality diameter. If chord passes through centre it is the diameter itself with length . So no chord exceeds . Also from formula with equality when .

23. Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.

Let chord through be with midpoint . Then is distance from centre to chord. For a fixed point on chord, with equality when (chord perpendicular to ). Since chord length decreases as increases, the smallest occurs when is maximal, i.e., . Hence the chord through perpendicular to has maximal distance and minimal length, and any other chord through is longer.

24. How would you use the following figure to justify the statement that the angle in a semicircle is 90°?

In Fig. 5.30, is diameter with centre , and are radii to point on circle. The central straight angle is subtended by diameter. The angle at circumference subtending same diameter is half the central angle by Theorem 9: . The figure shows isosceles triangles and with base angles and , so giving , which is the angle in semicircle.

25. In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C' D' is perpendicular to AB.

Since and are both perpendicular to diameter , they are parallel to each other. Their midpoints of and of lie on the line that is the perpendicular bisector of both chords, which passes through centre and is perpendicular to the chords. Because chords are perpendicular to , the line joining midpoints is collinear with 's perpendicular direction, hence is perpendicular to (in fact lies on a line perpendicular to through ). By symmetry of the circle with respect to , the midpoints are aligned vertically.

26. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?

In Fig. 5.31, centre and quadrilateral cyclic. Let central angles be for arcs . Each inscribed angle is half its opposite central angle: , etc., or directly reflex and . Summing gives . Similarly . The figure labels arcs whose sum is , halving gives opposite sums.