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GANITA MANJARI | Chapter 5 - Circles | Verbatim Q&A
CHAPTER 5

I'm Up and Down, and Round and Round

Verbatim Questions & Solutions | GANITA MANJARI | Class 9 Maths | Part I

Think and Reflect

Page 93
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?
Answer
Fold the paper along a diameter, then fold again along another diameter. The intersection point of the creases is the centre.

EXERCISE SET 5.1 (Page 98)

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?
Solution
∠C = 50°. All angles are acute, 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?
Solution
The triangle is obtuse-angled (∠A = 100° > 90°). So the circumcentre lies outside the triangle.
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.
Solution
OA = OB = OC = radius of the circumcircle. All three are equal (approximately 4.2 cm).
4.
What is the least possible radius of a circle through two points A and B?
Solution
The least possible radius is half the length of AB (when AB is the diameter).

Think and Reflect

Page 94
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?
2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
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?
Answer
1. Square: rotational symmetry of order 4, 4 lines of symmetry; Pentagon: order 5, 5 lines; Hexagon: order 6, 6 lines.
2. Longest chord = diameter = 10 units. Smallest chord can be arbitrarily small (approaching a point).
3. The locus of points equidistant from two given points is the perpendicular bisector of the segment joining them.
Page 95
1. How many circles pass through two points on a plane?
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?
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?
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?
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?
Answer
1. Infinitely many circles pass through two points.
2. Yes, circles of all radii greater than AB/2 exist. Smallest radius = AB/2. No largest radius (radius can be arbitrarily large).
3. The radius increases.
4. Less curved (the circle becomes flatter).
5. Infinitely many squares with A and B on the boundary. Two squares with A and B as corners (one on each side of AB).

Think, Draw and Infer

Page 98
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?
2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
Answer
1. No such point exists (unless the points coincide). Perpendicular bisectors of AB and BC are parallel. No circle can pass through three collinear points. A line cannot cut a circle at three distinct points (maximum two).
2. Yes, for any rotation of the triangle about the centre of the circumcircle, we get a congruent triangle inscribed in the same circle.

EXERCISE SET 5.2 (Page 100)

1.
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Solution
Let AB be a chord and O the centre. Then OA = OB (radii). So ΔOAB is isosceles with OA = OB.
2.
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Solution
Let the two triangles be OAB and OCD with AB = CD. Also OA = OB = OC = OD = radius. By SSS congruence, ΔOAB ≅ ΔOCD.

EXERCISE SET 5.3 (Page 101)

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?
Solution
Consider ΔCMA and ΔCMB. CM is common. ∠CMA = ∠CMB = 90°. CA = CB (radii). By RHS congruence, ΔCMA ≅ ΔCMB. Hence AM = BM.
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.
Solution
Since AB = AC, the perpendicular bisector of BC passes through A. But the perpendicular bisector of any chord passes through the centre. So the altitude from A to BC passes through the centre.
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.
Solution
For chord of length 6, distance from centre = √(5² - 3²) = 4 cm. For chord of length 8, distance = √(5² - 4²) = 3 cm. Since they are on opposite sides, distance between midpoints = 4 + 3 = 7 cm.

EXERCISE SET 5.4 (Page 104)

1.
Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.
Solution
For a chord of length l at distance d from centre, r² = d² + (l/2)². If two chords have same length, then (l/2)² is same, so d² = r² - (l/2)² is same. Hence d is same.
2.
Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
Solution
In right triangles AEC and GHC, CE = CH, AC = GC (radii). By RHS congruence, ΔAEC ≅ ΔGHC. So AE = GH. Hence AB = 2AE = 2GH = GF.
3.
Solve the previous question using the Baudhayana-Pythagoras theorem.
Solution
AB/2 = √(r² - CE²) and GF/2 = √(r² - CH²). Since CE = CH, AB/2 = GF/2, so AB = GF.

EXERCISE SET 5.5 (Pages 105-106)

1.
Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Solution
Length = 2√(7² - 6²) = 2√(49 - 36) = 2√13 cm.
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 2√(r² - d²).
Solution
Half the chord forms a right triangle with radius r and distance d. By Baudhayana-Pythagoras, (chord/2)² = r² - d², so chord = 2√(r² - d²).
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 = 2AB? Give reasons for your answer.
Solution
No. Let d₁ = distance of AB, d₂ = distance of CD. Given d₁ = 2d₂. Then AB = 2√(r² - (2d₂)²) and CD = 2√(r² - d₂²). The relationship is not simply CD = 2AB.

EXERCISE SET 5.6 (Pages 110-111)

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?
Solution
ΔOAB is isosceles with OA = OB = 12 and ∠AOB = 60°. So it is equilateral. Hence AB = 12 cm.
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?
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
(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?
Solution
(i) No, angles in the same segment are equal.
(ii) Yes, they lie on the same arc (same side of chord AB).
(iii) Yes, if they lie on the same side of AB and subtend equal angles, then A, B, X, Y are concyclic.
3.
Find x in Fig.5.26.
Solution
Without the figure, the answer depends on the specific angles shown. Typically, using the theorem that angles in the same segment are equal, x = given angle.

Think and Reflect

Page 113
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.
Answer
In a cyclic quadrilateral, opposite angles sum to 180°. Here ∠A + ∠C = 80° + 100° = 180° and ∠B + ∠D = 110° + 70° = 180°. So yes, such a quadrilateral can be drawn.

END-OF-CHAPTER EXERCISES (Pages 114-116)

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?
Solution
Length = 2√(13² - 5²) = 2√(169 - 25) = 2√144 = 2 × 12 = 24 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?
Solution
Angle at the circumference = half the angle at the centre = 35°.
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.
Solution
Radius = 13 cm. Distance = √(13² - 12²) = √(169 - 144) = √25 = 5 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?
Solution
Length = 2√(15² - 9²) = 2√(225 - 81) = 2√144 = 2 × 12 = 24 cm.
5.
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Solution
Any point on the perpendicular bisector of chord AB is equidistant from A and B. The centre O is equidistant from A and B (since OA = OB = radius). Hence O lies on the perpendicular bisector.
6.
The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB? Explain your reasoning.
Solution
∠ACB = 90° (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?
Solution
In a cyclic quadrilateral, opposite angles sum to 180°. So ∠C = 180° - 75° = 105°, ∠D = 180° - 110° = 70°.
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.
Solution
In a cyclic quadrilateral, ∠P + ∠R = 180°. So (2x + 10) + (3x - 20) = 180 ⇒ 5x - 10 = 180 ⇒ 5x = 190 ⇒ x = 38. Then ∠P = 2(38)+10 = 86°, ∠R = 3(38)-20 = 94°.
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.
Solution
Radius = √(6² + (16/2)²) = √(36 + 64) = √100 = 10 cm.
10.
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Solution
Using Brahmagupta's formula: s = (5+5+12+12)/2 = 17. Area = √[(17-5)(17-5)(17-12)(17-12)] = √[12 × 12 × 5 × 5] = √(144 × 25) = 12 × 5 = 60 sq units.
*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?
Solution
If the quadrilateral is acute (all angles < 90°?), centre is inside. If one angle > 90°, centre is outside. Alternatively, check if sum of opposite angles > 180°.
*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.
Solution
Let chords AB and CD intersect at O. If AB = CD, then using properties of intersecting chords, OA × OB = OC × OD. With additional conditions, equality of segments can be shown.
*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?)
Solution
Construct a right triangle with legs 3 cm and 3 cm (half-chord), then hypotenuse = √(3²+3²) = √18 = 3√2 ≈ 4.24 cm as radius. Then draw circle.
*14.
Show that rectangle is the only parallelogram that can be inscribed in a circle.
Solution
In a cyclic quadrilateral, opposite angles sum to 180°. In a parallelogram, opposite angles are equal. So each angle must be 90°. Hence it is a rectangle.
*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.
Solution
In a rectangle, diagonals bisect each other. The centre of the circumcircle is equidistant from all vertices, so it is the midpoint of the diagonals.
*16.
Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Solution
The midpoints lie on a circle concentric with the given circle.
*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".
Solution
Since AB = AC, they are equidistant from O. O lies on the perpendicular bisector of BC. Also, O is equidistant from AB and AC, so O lies on the angle bisector of ∠BAC.
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.
Solution
Let distances from centre be d₁ and d₂. Then 2√(r² - d₁²) = 10, 2√(r² - d₂²) = 24, and |d₁ - d₂| = 7. Solving gives r = 13 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.
Solution
Side length = r. Distance from centre = r√3/2.
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.
Solution
∠MOP is the central angle subtended by arc MP. ∠MNP is the angle at the circumference subtended by the same arc. So ∠MOP = 2∠MNP.
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).
Solution
∠CDE is supplementary to ∠CDA. In cyclic quadrilateral, ∠CDA + ∠ABC = 180°. So ∠CDE = 180° - ∠CDA = ∠ABC.
*22.
"There is no chord of a circle that is longer than its diameter." How do you justify this statement?
Solution
The maximum distance between two points on a circle is the diameter. Any chord is ≤ diameter.
*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.
Solution
For a fixed distance d from O, chord length = 2√(r² - d²). To minimise chord length, maximise d. The maximum d is OA, achieved when chord is perpendicular to OA.
24.
How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Solution
In the figure, the central angle is 180°, so the inscribed angle is half of that, i.e., 90°.
*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.
Solution
By symmetry and properties of perpendicular bisectors, MM' is parallel to AB or perpendicular depending on configuration.
*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°?
Solution
The inscribed angles subtend arcs whose sum is 360°, leading to sum of opposite angles = 180°.

Chapter Summary (Pages 116-117)

  • A circle is the set of all points in a plane that lie at a given distance (the radius) from a fixed point called its centre.
  • A circle has reflection symmetry across any diameter and rotational symmetry about its centre.
  • Infinitely many circles can be drawn through two given points; their centres lie on the perpendicular bisector.
  • Given any three non-collinear points, a unique circle (circumcircle) can be drawn through them.
  • Equal chords subtend equal angles at the centre, and vice versa.
  • The perpendicular from the centre to a chord bisects the chord, and vice versa.
  • Chords of equal length are equidistant from the centre, and vice versa.
  • The longer chord is closer to the centre.
  • The angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.
  • The angle subtended by a diameter at any point on the circle is 90°.
  • In a cyclic quadrilateral, the sum of opposite angles is 180°.

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