# Circles

Chapters

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- Solutions
- PDF Download
- Question Papers
- Videos

Solutions

Name the following from the given figure where ’O’ is the centre of the circle.

State true or false.

- i) A circle divides the plane on which it lies into three parts. ( )
- ii) The area enclosed by a chord and the minor arc is minor segment. ( )
- iii) The area enclosed by a chord and the major arc is major segment. ( )
- iv) A diameter divides the circle into two unequal parts. ( )
- v) A sector is the area enclosed by two radii and a chord. ( )
- vi) The longest of all chords of a circle is called a diameter. ( )
- vii) The mid point of any diameter of a circle is the centre. ( )

Solution:

- i) True
- ii) True
- iii) True
- iv) False
- v) False
- vi)True
- vii) True

In the figure, if AB = CD and ∠AOB = 90° find ∠COD.

Solution:

‘O’ is the centre of the circle.

AB = CD (equal chords from the figure)

∴ ∠AOB = ∠COD

[ ∵ equal chords make equal angles at the centre]

∴ ∠COD = 90° [ ∵ ∠AOB = 90° given]

In the figure, PQ = RS and ∠ORS = 48°.

Find ∠OPQ and ∠ROS.

Solution:

‘O’ is the centre of the circle.

PQ = RS [given, equal chords]

∴∠POQ = ∠ROS [ ∵ equal chords make equal angles at the centre]

∴ In ΔROS

∠ORS + ∠OSR + ∠ROS = 180°

[angle sum property]

∴ 48° + 48° + ∠ROS = 180°

[ ∵ OR = OS(radii); ΔORS is isosceles]

∴ ∠ROS = 180° - 96° = 84°

Also ∠POQ = ∠ROS = 84°

∴ ∠OPQ = ∠OQP

[∵ OP = OQ; radii]

=1/2 [180°-84°] = 48°

In the figure, PR and QS are two diameters. Is PQ = RS ?

Solution:

‘O’ is the centre of the circle.

[ ∵ PR, QS are diameters]

OP = OR (∵ radii) .

OQ = OS (∵ radii)

∠POQ = ∠ROS [vertically opp. angles]

∴ ΔOPQ ≅ ΔORS [SAS congruence]

∴ PQ = RS [CPCT]

Draw the following triangles and construct circumcircles for them.

(OR)

In ΔABC, AB = 6 cm, BC = 7 cm and ∠A = 60°.

Construct a circumcircle to the triangle XYZ given XY = 6cm, YZ = 7cm and ∠Y = 60°. Also, write steps of construction.

Solution:

Steps of construction :

- Draw the triangle with given mea-sures.
- Draw perpendicular bisectors to the sides.
- The point of concurrence of per-pendicular bisectors be S’.
- With centre S; SA as radius, draw a circle which also passes through B and C.
- This is the required circumcircle.

ii) In ΔPQR; PQ = 5 cm, QR = 6 cm and RP = 8.2 cm.

Steps of construction:

- Draw ΔPQR with given measures.
- Draw perpendicular to PQ, QR and RS; let they meet at ‘S’.
- With S as centre and SP as radius draw a circle.
- This is the required circumcircle.

iii) In ΔXYZ, XY = 4.8 cm, ∠X = 60° and ∠Y = 70°.

Steps of construction:

- Draw ΔXYZ with given measures.
- Draw perpendicular bisectors to the sides of ΔXYZ, let the point of con-currence be S’.
- Draw the circle (S, SX).
- This is the required circumcircle.

Draw two circles passing A, B where AB = 5.4 cm.

(OR)

Draw a line segment AB with 5.4 cm. length and draw two different circles that passes through both A and B.

Solution:

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.

Join A, B to form the common chord

AB Let ‘O’ be the midpoint of AB.

Join ‘O’ with P and Q.

Now in 8APO and ΔBPO

AP = BP (radii)

PO = PO (common)

AO = BO (∵ O is the midpoint)

∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)

Also ∠AOP = ∠BOP (CPCT)

But these are linear pair of angles.

∴ ∠AOP = ∠BOP = 90°

Similarly in ΔAOQ and ΔBOQ

AQ = BQ (radii)

AO = BO (∵ O is the midpoint of AB)

OQ = OQ (common)

∴ AAOQ ≅ ABOQ

Also ∠AOQ = ∠BOQ (CPCT)

Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)

∴ ∠AOQ = ∠BOQ =180°/2= 90°

Now ∠AOP + ∠AOQ = 180°

∴ PQ is a line.

Hence the proof.

If two intersecting chords of a circle make equal angles with diameter pass¬ing through their point of intersection, prove that the chords are equal.

∠AEO = ∠DEO

Drop two perpendiculars OL and OM from ‘O’ to AB and CD;

Now in ΔLEO and ΔMEO

∠LEO = ∠MEO [given]

EO = EO [Common]

∠ELO = ∠EMO [construction 90°]

∴ ΔLEO ≅ ΔMEO

[ ∵ A.A.S. congruence]

∴ OL = OM [CPCT]

i.e., The two chords AB and CD are at equidistant from the centre ‘O’.

∴ AB = CD

[∵ Chords which are equi-distant from the centre are equal]

Hence proved.

In the given figure, AB is a chord of circle with centre ‘O’. CD is the diam-eter perpendicular to AB. Show that AD = BD.

Solution:

CD is diameter, O is the centre.

CD ⊥ AB; Let M be the point of inter-section.

Now in ΔAMD and ΔBMD

AM = BM [ ∵ radius perpendicular to a chord bisects it]

∠AMD =∠BMD [given]

DM = DM (common)

∴ ΔAMD ≅ ΔBMD

⇒ AD = BD [C.P.C.T]

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:

’O’ is the centre

∠AOB = 100°

In the figure, ∠BAD = 40° then find ∠BCD.

Solution:

‘O’ is the centre of the circle.

∴ In ΔOAB; OA = OB (radii)

∴ ∠OAB = ∠OBA = 40°

(∵ angles opp. to equal sides)

Now ∠AOB = 180° - (40° + 40°)

(∵ angle sum property of ΔOAB)

= 180°-80° = 100°

But ∠AOB = ∠COD = 100°

Also ∠OCD = ∠ODC [OC = OD]

= 40° as in ΔOAB

∴ ∠BCD = 40°

(OR)

In ΔOAB and ΔOCD

OA = OD (radii)

OB = OC (radii)

∠AOB = ∠COD (vertically opp. angles)

∴ ΔOAB ≅ ΔOCD

∴ ∠BCD = ∠OBA = 40°

[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

In the figure, ‘O’ is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.

If a parallelogram is cyclic, then it is a rectangle. Justify.

Solution:

Let □ABCD be a parallelogram such

that A, B, C and D lie on the circle.

∴∠A + ∠C = 180° and ∠B + ∠D = 180°

[Opp. angles of a cyclic quadri lateral are supplementary]

But ∠A = ∠C and ∠B = ∠D

[∵ Opp. angles of a ‖gm are equal]

∴∠A = ∠C =∠B =∠D =180/2= 90°

Hence □ABCD is a rectangle

In the figure, ‘O’ is the centr of the circle. 0M = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.

OM bisects AB.

∴ AM =AB/2=8/2= 4 cm

OA^{2}= OM^{2}+ AM^{2}[ ∵ Pythagoras theorem]

OA =√3^{2}+4^{2}

=√9+16=√25

= 5cm

In the figure, ‘O’ is the centre of the circle and OM, ON are the perpen-diculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6 cm. Find RS.

Solution:

‘O’ is the centre of the circle.

OM = ON and 0M ⊥ PQ; ON ⊥ RS

Thus the chords FQ and RS are equal.

[ ∵ chords which are equidistant from the centre are equal in length]

∴ RS = PQ = 6cm

A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the

radius of the circle.

Solution:

A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle.

∴ Radius = 4 cm.

Draw a circle with any radius and then draw two chords equidistant

from the centre.

Solution:

- Draw a circle with centre P.
- Draw any two radii.
- Mark off two points M and N oh these radii. Such that PM = PN.
- Draw perpendicular through M and N to these radii.

In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD.

Solution:

‘O’ is the centre of the circle.

AB, CD are equal chords

⇒ They subtend equal angles at the centre.

∴ ∠AOB =∠COD = 70°

Now in ΔOCD

∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]

∴ ∠OCD + ∠ODC + 70° = 180°

= ∠OCD +∠ODC = 180° – 70° = 110°

∴ ∠OCD + ∠ODC = 110° = 55°

Find the values of x and y in the figures given below.

i)

Solution:

From the figure x = y [∵ angles opp. to opp. to equal sides]

But x + y + 30° = 180°

∴ x + y = 180° - 30° = 150°

⇒ x + y = 150°/2= 75°

ii)

Solution:

From the figure x° + 110° = 180°

[ ∵ Opp. angles of a cyclic quad, are supplementary]

y + 85°= 180°

∴ x= 180° - 110°; y = 180° – 85°

x = 70°; y = 95°

iii)

Solution:

From the figure x = 90° [angle in a semi-circle]

∴ y = 90° – 50° [∵ angle sum property]

= 40°

Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle.

Solution:

Given : ∠A + ∠C = 180°

∴ ∠B + ∠D = 360° – 180°

[ ∵ sum of the four angles of a quad. is 360 ].

Now in □ABCD, sum of the pairs of opp. angles is 180°.

∴ □ABCD must be a cyclic quadrilateral, i.e., D also lie on the same circle on which the vertices A, B and C lie. Hence proved.

Prove that a cyclic rhombus is a square.

Solution:

Let □ABCD be a cyclic rhombus,

i.e., AB = BC = CD = DA and

∠A + ∠C = ∠B + ∠D = 180°

But a rhombus is basically a parallelo-gram.

For each of the following, draw a circle and inscribe the figure given. If a poly¬gon of the given type can’t be in-scribed, write not possible.

- a) Rectangle
- b) Trapezium .
- c) Obtuse triangle
- d) Non-rectangular parallelogram
- e) Acute isosceles triangle
- f) A quadrilateral PQRS with PR as diameter.

Solution:

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