NCERT Solutions Class 9 Maths Chapter 9: Circles

(i) The centre of a circle lies in interior of the circle.

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

(iii) The longest chord of a circle is a diameter of the circle.

(iv) An arc is a semicircle when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and chord of the circle.

(vi) A circle divides the plane, on which it lies, in 3 (three) parts.

 

(i) True. Any line segment drawn from the centre of the circle to any point on it is the radius of the circle and will be of equal length.

(ii) False. There can be infinite numbers of equal chords in a circle.

(iii) False. For unequal arcs, there can be major and minor arcs. So, equal arcs on a circle cannot be said to be major arcs or minor arcs.

(iv) True. Any chord whose length is twice as long as the radius of the circle always passes through the centre of the circle, and thus, it is known as the diameter of the circle.

(v) False. A sector is a region of a circle between the arc and the two radii of the circle.

(vi) True. A circle is a 2d figure, and it can be drawn on a plane.

 

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).

 

Consider the following diagram.

 

Here, it is given that ∠AOB = ∠COD, i.e., they are equal angles.

Now, we will have to prove 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.)

 

In these two circles, no point is common.

 

Here, only one point, ‘P’, is common.

 

Even here, P is the common point.

 

Here, two points are common, which are P and Q.

 

No point is common in the above circle.

 

The construction steps to find the centre of the circle is:

Step I: Draw a circle first.

Step II: Draw 2 chords, AB and CD, in the circle.

Step III: Draw the perpendicular bisectors of AB and CD.

Step IV: Connect the two perpendicular bisectors at a point. This intersection point of the two perpendicular bisectors is the centre of the circle.

 

It is given that two circles intersect each other at P and Q.

To prove:

OO’ is perpendicular bisector of PQ.

(i) PR = RQ

(ii) ∠PRO = ∠PRO’ = ∠QRO = ∠QRO’ = 900

Proof:

In triangles ΔPOO’ and ΔQOO’,

OP = OQ and O’P = O’Q (Since they are also the radii.)

OO’ = OO’ (It is the common side.)

So, it can be said that ΔPOO’ ≅ ΔQOO’ (SSS Congruence rule)

∴ ∠POO’ = ∠QOO’ (c.p.c.t)— (i)

Even triangles ΔPOR and ΔQOR are similar by SAS congruency.

OP = OQ (Radii)

∠POR = ∠QOR (As ∠POO’ = ∠QOO’)

OR = OR (Common arm)

So, ΔOPO’ ≅ ΔOQO’ (SAS Congruence rule)

∴ PR = QR and ∠PRO = ∠QRO (c.p.c.t) …. (ii)

As PQ is a line

∠PRO + ∠QRO = 180°

∠PRO + ∠PRO = 180° (Using (ii))

2∠PRO = 180°

∠PRO = 90°

So ∠QRO = ∠PRO = 90°

Here,

∠PRO’ = ∠QRO = 90° and ∠QRO’ = ∠PRO = 90° (Vertically opposite angles)

∠PRO = ∠QRO = ∠PRO’ = ∠QRO’ = 90°

So, OO’ is the perpendicular bisector of PQ.

 

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.

 

 

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)

 

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)

 

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

 

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

OA2 = OM2 +AM2

⇒ 52 = x2 +y2 — (i)

Again, by applying the Pythagorean theorem in ΔAMB,

AB2 = BM2 +AM2

⇒ 62 = (5-x)2+y2 — (ii)

Subtracting equation (i) from equation (ii), we get

36-25 = (5-x)2 +y2 -x2-y2

Now, solving this equation, we get the value of x as

x = 7/5

Substituting the value of x in equation (i), we get

y2 +(49/25) = 25

⇒ y2 = 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.

 

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,

AB2 = BD2+AD2

⇒ AD2 = AB2 -BD2

⇒ AD2 = a2 -(a/2)2

⇒ AD2 = 3a2/4

⇒ 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.

 

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.

 

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°

 

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.

 

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°

 

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°

 

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°

 

 

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°

 

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.

 

Given:

A trapezium ABCD with $AB\parallel CD$ and $AD=BC$.

To prove:

The trapezium is cyclic, meaning the opposite angles of the trapezium add up to 180∘180^\circ180∘.

Proof:

1. Draw the trapezium: Let $ABCD$ be the trapezium where $AB\parallel CD$ and $AD=BC$.

2. Consider the angles: In any trapezium with one pair of parallel sides, the sum of the interior angles on the same side of the transversal (non-parallel sides) is supplementary.

   This means:

   ∠A+∠D=180∘(since AB∥CD and AD is the transversal).\angle A + \angle D = 180^\circ \quad \text{(since \( AB \parallel CD \) and \( AD \) is the transversal)}.∠A+∠D=180∘(since AB∥CD and AD is the transversal).

   Similarly:

   ∠B+∠C=180∘(since AB∥CD and BC is the transversal).\angle B + \angle C = 180^\circ \quad \text{(since \( AB \parallel CD \) and \( BC \) is the transversal)}.∠B+∠C=180∘(since AB∥CD and BC is the transversal).

3. Using the fact that $AD=BC$: Since the non-parallel sides $AD$ and $BC$ are equal, the trapezium is isosceles. Therefore:

   $$\angle A=\angle B\text{and}\angle D=\angle C.$$

4. Substitute into the equation: From step 2, we know:

   ∠A+∠D=180∘.\angle A + \angle D = 180^\circ.∠A+∠D=180∘.

   Since $\angle A=\angle B$ and $\angle D=\angle C$, it follows that:

   ∠B+∠C=180∘.\angle B + \angle C = 180^\circ.∠B+∠C=180∘.

5. Conclusion: Since the sum of the opposite angles ∠A+∠D=180∘\angle A + \angle D = 180^\circ∠A+∠D=180∘ and ∠B+∠C=180∘\angle B + \angle C = 180^\circ∠B+∠C=180∘, the trapezium satisfies the property of a cyclic quadrilateral. Hence, the trapezium can be inscribed in a circle.

Thus, the trapezium is cyclic. Q.E.D.\boxed{\text{Q.E.D.}}Q.E.D.​

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

 

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.

 

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

 

Consider the following diagram.

 

In ΔPOO’ and ΔQOO’

OP = OQ          (Radius of circle 1)

O’P = O’Q        (Radius of circle 2)

OO’ = OO’        (Common arm)

So, by SSS congruency, ΔPOO’ ≅ ΔQOO’

Thus, ∠OPO’ = ∠OQO’ (proved).

 

Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.

We know that AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2 = 5/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6−x.

Consider ΔMOB,

OB2 = OM2+MB2

Or,

 

Consider ΔNOD,

OD2 = ON2 + ND2

Or

 

We know, OB = OD (radii)

From equation 1 and equation 2, we get

 

Now, from equation (2), we have,

OD2= 12 +(121/4)

Or OD = (5/2)×√5 cm

 

Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.

We know that AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2 = 5/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6−x.

Consider ΔMOB,

OB2 = OM2+MB2

Or,

 

Consider ΔNOD,

OD2 = ON2 + ND2

Or

 

We know, OB = OD (radii)

From equation 1 and equation 2, we get

 

Now, from equation (2), we have,

OD2= 12 +(121/4)

Or OD = (5/2)×√5 cm

 

Consider the following diagram.

 

Here, AB and CD are 2 parallel chords. Now, join OB and OD.

Distance of smaller chord AB from the centre of the circle = 4 cm

So, OM = 4 cm

MB = AB/2 = 3 cm

Consider ΔOMB.

OB2 = OM2+MB2

Or, OB = 5cm

Now, consider ΔOND.

OB = OD = 5 (Since they are the radii.)

ND = CD/2 = 4 cm

Now, OD2= ON2+ND2

Or, ON = 3 cm

 

Consider the diagram.

 

Here AD = CE

We know any exterior angle of a triangle is equal to the sum of interior opposite angles.

So,

∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)

DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

So,

∠DAE = (½)∠DOE ——————-(ii)

Similarly, ∠AEC = (½)∠AOC  ——————-(iii)

Now, from equations (i), (ii), and (iii), we get

(½)∠DOE = ∠ABC+(½)∠AOC

Or, ∠ABC = (½)[∠DOE-∠AOC]  (Hence, proved)

 

To prove: A circle drawn with Q as the centre will pass through A, B and O (i.e., QA = QB = QO).

Since all sides of a rhombus are equal,

AB = DC

Now, multiply (½) on both sides.

(½)AB = (½)DC

So, AQ = DP

BQ = DP

Since Q is the midpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO, we get

QA = QB = QO (Hence, proved)

 

Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.

So, ∠AEC+∠CBA = 180°

As ∠AEC and ∠AED are linear pairs,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … (1)

We know in a parallelogram, opposite angles are equal.

So, ∠ADE = ∠CBA … (2)

Now, from equations (1) and (2), we get

∠AED = ∠ADE

Now, AD and AE are angles opposite to equal sides of a triangle.

∴ AD = AE (proved)

 

Here, chords AB and CD intersect each other at O.

Consider ΔAOB and ΔCOD.

∠AOB = ∠COD (They are vertically opposite angles.)

OB = OD (Given in the question.)

OA = OC (Given in the question.)

So, by SAS congruency, ΔAOB ≅ ΔCOD

Also, AB = CD (By CPCT)

Similarly, ΔAOD ≅ ΔCOB

Or, AD = CB (By CPCT)

In quadrilateral ACBD, opposite sides are equal.

So, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

So, ∠A = ∠C

Also, as ABCD is a cyclic quadrilateral,

∠A+∠C = 180°

⇒∠A+∠A = 180°

Or, ∠A = 90°

As ACBD is a parallelogram and one of its interior angles is 90°, so, it is a rectangle.

∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.

 

 

Consider the following diagram.

 

Here, ABC is inscribed in a circle with centre O, and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F, respectively.

Now, join DE, EF and FD.

As angles in the same segment are equal, so,

∠EDA = ∠FCA ————-(i)

∠FDA = ∠EBA ————-(i)

By adding equations (i) and (ii), we get

∠FDA+∠EDA = ∠FCA+∠EBA

Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B

We know, ∠A +∠B+∠C = 180°

So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]

∠FDE = [90-(∠A/2)]

In a similar way,

∠FED = [90° -(∠B/2)] °

And,

∠EFD = [90° -(∠C/2)] °

 

The diagram will be

 

Here, ∠APB = ∠AQB (as AB is the common chord in both the congruent circles.)

Now, consider ΔBPQ.

∠APB = ∠AQB

So, the angles are opposite to equal sides of a triangle.

∴ BQ = BP

 

Consider this diagram.

 

Here, join BE and CE.

Now, since AE is the bisector of ∠BAC,

∠BAE = ∠CAE

Also,

∴arc BE = arc EC

This implies chord BE = chord EC

Now, consider triangles ΔBDE and ΔCDE.

DE = DE     (It is the common side)

BD = CD     (It is given in the question)

BE = CE      (Already proved)

So, by SSS congruency, ΔBDE ≅ ΔCDE.

Thus, ∴∠BDE = ∠CDE

We know, ∠BDE = ∠CDE = 180°

Or, ∠BDE = ∠CDE = 90°

∴ DE ⊥ BC (Hence, proved).