NCERT Class 9 Circles Solutions Chapter 9 PDF Download

NCERT Solutions Class 9 Maths Chapter 9 Circles Solutions play an essential role in helping students understand the fundamental concepts of circle geometry. These solutions cover all important topics introduced in Class 9 Maths Circles.

It includes definitions, properties, theorems, and problem-solving techniques. By studying these Circles Class 9 NCERT Solutions, you can develop a strong grasp of concepts such as the definition of a circle, radius, diameter, chords, arcs, tangents, angles formed in a circle, and theorems.

NCERT Solutions Class 9 Chapter 9 Circles Overview

Circles Class 9 NCERT Solutions help students simplify complex geometric problems by breaking them down into easy-to-follow steps. These explanations build conceptual clarity and help students understand how different geometric elements are.

By practicing Class 9 Maths Circles​ solutions regularly, you can gain conceptual clarity in circle geometry. It will also help improve accuracy and speed in solving questions. 

NCERT Class 9 Circle Solution

Here are  the Circles Class 9 NCERT Solutions given to help you prepare Chapter 9 Circles thoroughly: 

NCERT Class 9 Circles Exercise 9.1 (Page No: 118)

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Solution:

To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both circles is equal from the centre.For the second part of the question, it is given that AB = CD, i.e., two equal chords. Now, it is to be proven that angle AOB is equal to angle COD.

Proof:

Consider the triangles ΔAOB and ΔCOD. OA = OC and OB = OD (Since they are the radii of the circle.) AB = CD (As given in the question.) So, by SSS congruency, ΔAOB ≅ ΔCOD ∴ By CPCT, we have, ∠ AOB = ∠ COD (Hence, proved).

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Solution:

Consider the following diagram.

Here, it is given that ∠ AOB = ∠ COD, i.e., they are equal angles. Now, it proves that the line segments AB and CD are equal, i.e., AB = CD.

Proof:

In triangles AOB and COD, ∠ AOB = ∠ COD (As given in the question.) OA = OC and OB = OD (These are the radii of the circle.) So, by SAS congruency, ΔAOB ≅ ΔCOD ∴ By the rule of CPCT, we have, AB = CD (Hence, proved.)

NCERT Class 9 Circle Exercise 9.2 (Page No: 122)

1. Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.

Solution:

The perpendicular bisector of the common chord passes through the centres of both circles. As the circles intersect at two points, we can construct the above figure. Consider AB as the common chord and O and O’ as the centres of the circles. 

O’A = 5 cm 

OA = 3 cm 

OO’ = 4 cm [Distance between centres is 4 cm.] 

As the radius of the bigger circle is more than the distance between the two centres, we know that the centre of the smaller circle lies inside the bigger circle. The perpendicular bisector of AB is OO’. OA = OB = 3 cm As O is the midpoint of AB AB = 3 cm + 3 cm = 6 cm The length of the common chord is 6 cm. It is clear that the common chord is the diameter of the smaller circle.

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Let AB and CD be two equal cords (i.e., AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E. It is now to be proven that the line segments AE = DE and CE = BE

Construction Steps

Step 1: From the centre of the circle, draw a perpendicular to AB, i.e., OM ⊥ AB.

Step 2: Similarly, draw ON ⊥ CD.

Step 3: Join OE.

Now, the diagram is as follows:

Proof:

From the diagram, it is seen that OM bisects AB, and so OM ⊥ AB Similarly, ON bisects CD, and so ON ⊥ CD. It is known that AB = CD. So, AM = ND —

(i) and MB = CN — 

(ii) Now, triangles ΔOME and ΔONE are similar by RHS congruency, since ∠ OME = ∠ ONE (They are perpendiculars.) OE = OE (It is the common side.) OM = ON (AB and CD are equal, and so they are equidistant from the centre.) 

∴ ΔOME ≅ ΔONE ME = EN (by CPCT) — 

(iii) Now, from equations (i) and (ii), we get AM+ME = ND+EN So, AE = ED Now from equations (ii) and (iii), we get MB-ME = CN-EN So, EB = CE (Hence, proved)

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:

From the question, we know the following: (i) AB and CD are 2 chords which are intersecting at point E. (ii) PQ is the diameter of the circle. (iii) AB = CD. Now, we will have to prove that ∠ BEQ = ∠ CEQ For this, the following construction has to be done.

Construction:

Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:

Now, consider the triangles ΔOEM and ΔOEN. Here, 

(i) OM = ON [The equal chords are always equidistant from the centre.] 

(ii) OE = OE [It is the common side.] 

(iii) ∠ OME = ∠ ONE [These are the perpendiculars.] 

So, by RHS congruency criterion, ΔOEM ≅ ΔOEN. Hence, by the CPCT rule, ∠ MEO = ∠ NEO ∴ ∠ BEQ = ∠ CEQ (Hence, proved)

4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 9.12).

Solution:

The given image is as follows: 

First, draw a line segment from O to AD, such that OM ⊥ AD. So, now OM is bisecting AD since OM ⊥ AD. Therefore, AM = MD — 

(i) Also, since OM ⊥ BC, OM bisects BC. Therefore, BM = MC — 

(ii) From equation (i) and equation (ii), AM-BM = MD-MC ∴ AB = CD

5. Three girls, Reshma, Salma and Mandip, are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Solution:

Let the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively. From the question, we know that AB = BC = 6cm So, the radius of the circle, i.e., OA = 5cm. 

Now, draw a perpendicular BM ⊥ AC. Since AB = BC, ABC can be considered an isosceles triangle. M is the mid-point of AC. BM is the perpendicular bisector of AC, and thus it passes through the centre of the circle. Now, let AM = y and OM = x So, BM will be = (5-x). 

By applying the Pythagorean theorem in ΔOAM, we get OA 2 = OM 2 +AM 2 ⇒ 5 2 = x 2 +y 2 — 

(i) Again, by applying the Pythagorean theorem in ΔAMB, AB 2 = BM 2 +AM 2 ⇒ 6 2 = (5-x) 2 +y 2 — 

(ii) Subtracting equation (i) from equation (ii), we get 36-25 = (5-x) 2 +y 2 -x 2 -y 2 Now, solving this equation, we get the value of x as x = 7/5

Substituting the value of x in equation (i), we get y 2 +(49/25) = 25 ⇒ y 2 = 25 – (49/25) Solving it, we get the value of y as y = 24/5 Thus, AC = 2×AM = 2×y = 2×(24/5) m AC = 9.6 m So, the distance between Reshma and Mandip is 9.6 m.

6. A circular park of radius 20m is situated in a colony. Three boys, Ankur, Syed and David, are sitting at equal distances on its boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Solution:

First, draw a diagram according to the given statements. The diagram will look as follows:

Here, the positions of Ankur, Syed and David are represented as A, B and C, respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle. AD ⊥ BC is drawn. 

Now, AD is the median of ΔABC, and it passes through the centre O. Also, O is the centroid of the ΔABC. OA is the radius of the triangle. OA = 2/3 AD Let the side of a triangle a metres, then BD = a/2 m. 

Applying Pythagoras’ theorem in ΔABD, AB 2 = BD 2 +AD 2 ⇒ AD 2 = AB 2 -BD 2 ⇒ AD 2 = a 2 -(a/2) 2 ⇒ AD 2 = 3a 2 /4 ⇒ AD = √3a/2 OA = 2/3 AD 20 m = 2/3 × √3a/2 a = 20√3 m So, the length of the string of the toy is 20√3 m.

Circles Class 9 Solutions Chapter 9 Exercise 9.3 (Page No: 127)

1. In Fig. 9.23, A, B and C are three points on a circle with centre O, such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:

It is given that, ∠ AOC = ∠ AOB+ ∠ BOC So, ∠ AOC = 60°+30° ∴ ∠ AOC = 90° It is known that an angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle. So, ∠ ADC = (½) ∠ AOC = (½)× 90° = 45°

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

Here, the chord AB is equal to the radius of the circle. In the above diagram, OA and OB are the two radii of the circle. Now, consider the ΔOAB. 

Here, AB = OA = OB = radius of the circle So, it can be said that ΔOAB has all equal sides, and thus, it is an equilateral triangle. 

∴ ∠ AOC = 60° And, ∠ ACB = ½ ∠ AOB So, ∠ ACB = ½ × 60° = 30° 

Now, since ACBD is a cyclic quadrilateral, ∠ ADB + ∠ ACB = 180° (They are the opposite angles of a cyclic quadrilateral) So, ∠ ADB = 180°-30° = 150° So, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 150° and 30°, respectively.

3. In Fig. 9.24, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Solution:

Since the angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle. So, the reflex ∠ POR = 2× ∠ PQR We know the values of angle PQR as 100°. 

So, ∠ POR = 2×100° = 200° 

∴ ∠ POR = 360°-200° = 160° 

Now, in ΔOPR, OP and OR are the radii of the circle. So, OP = OR Also, 

∠ OPR = ∠ ORP Now, we know the sum of the angles in a triangle is equal to 180 degrees. 

So, ∠ POR+ ∠ OPR+ ∠ ORP = 180° 

∠ OPR+ ∠ OPR = 180°-160° 

As ∠ OPR = ∠ ORP 2 ∠ OPR = 20°

Thus, ∠ OPR = 10°

4. In Fig. 9.25, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.

Solution:

We know that angles in the segment of the circle are equal, so, ∠BAC = ∠BDC Now. in the ΔABC, the sum of all the interior angles will be 180°. 

So, ∠ABC+∠BAC+∠ACB = 180° Now, by putting the values, ∠BAC = 180°-69°-31° So, ∠BAC = 80° ∴ ∠BDC = 80°

5. In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E, such that ∠ BEC = 130° and ∠ ECD = 20°. Find BAC.

Solution:

We know that the angles in the segment of the circle are equal. So, ∠ BAC = ∠ CDE Now, by using the exterior angles property of the triangle, In ΔCDE, we get ∠ CEB = ∠ CDE+∠ DCE We know that ∠ DCE is equal to 20°. So, ∠ CDE = 110° ∠ BAC and ∠ CDE are equal ∴ ∠ BAC = 110°

6. ABCD is a cyclic quadrilateral whose diagonals intersect at point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Solution:

Consider the following diagram.

Consider the chord CD. We know that angles in the same segment are equal. So, ∠ CBD = ∠ CAD 

∴ ∠ CAD = 70° 

Now, ∠ BAD will be equal to the sum of angles BAC and CAD. So, ∠ BAD = ∠ BAC+∠ CAD = 30°+70° 

∴ ∠ BAD = 100° 

We know that the opposite angles of a cyclic quadrilateral sum up to 180 degrees. So, ∠ BCD+∠ BAD = 180° It is known that ∠ BAD = 100° So, ∠ BCD = 80° Now, consider the ΔABC. 

Here, it is given that AB = BC Also, ∠ BCA = ∠ CAB (They are the angles opposite to equal sides of a triangle) ∠ BCA = 30° also, ∠ BCD = 80° ∠ BCA +∠ ACD = 80° Thus, ∠ ACD = 50° and ∠ ECD = 50°

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Draw a cyclic quadrilateral ABCD inside a circle with centre O, such that its diagonal AC and BD are two diameters of the circle.We know that the angles in the semi-circle are equal. So, ∠ ABC = ∠ BCD = ∠ CDA = ∠ DAB = 90° So, as each internal angle is 90°, it can be said that the quadrilateral ABCD is a rectangle.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

9. Two circles intersect at two points, B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q, respectively (see Fig. 9.27). Prove that ∠ ACP = ∠ QCD.

Solution:

Construction:

Join the chords AP and DQ. For chord AP, we know that angles in the same segment are equal. So, ∠ PBA = ∠ ACP — (i) Similarly, for chord DQ, ∠ DBQ = ∠ QCD — (ii) It is known that ABD and PBQ are two line segments which are intersecting at B. At B, the vertically opposite angles will be equal. ∴ ∠ PBA = ∠ DBQ — (iii) From equation (i), equation (ii) and equation (iii), we get ∠ ACP = ∠ QCD

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.

Solution:

First, draw a triangle ABC and then two circles having diameters of AB and AC, respectively. We will have to now prove that D lies on BC and BDC is a straight line.

Proof:

We know that angles in the semi-circle are equal. So, ∠ ADB = ∠ ADC = 90° Hence, ∠ ADB+∠ ADC = 180° 

∴ ∠ BDC is a straight line. So, it can be said that D lies on the line BC.

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠CBD.

Solution:

We know that AC is the common hypotenuse and ∠ B = ∠ D = 90°. Now, it has to be proven that ∠ CAD = ∠ CBD

Since ∠ ABC and ∠ ADC are 90°, it can be said that they lie in a semi-circle. So, triangles ABC and ADC are in the semi-circle, and the points A, B, C and D are concyclic. Hence, CD is the chord of the circle with centre O. We know that the angles which are in the same segment of the circle are equal. ∴ ∠ CAD = ∠ CBD

12. Prove that a cyclic parallelogram is a rectangle.

Solution:

It is given that ABCD is a cyclic parallelogram, and we will have to prove that ABCD is a rectangle.

Proof:

Thus, ABCD is a rectangle.

NCERT Solutions Class 9 MathsChapter 9 Circles PDF

The NCERT Class 9 Circle Solution​ PDF provides comprehensive, step-by-step answers to all textbook exercises. This PDF covers all important circle concepts such as chords, tangents, arcs, angles, and essential theorems. You can easily revise the chapter, clear doubts, and strengthen their understanding by referring to well-structured solutions.

NCERT Solutions for Class 9 Maths Chapter 9 PDF

CIRCLES in 1 Shot: Full Chapter

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