# Circles

Chapters

- cbse-ncert-9th-class-Chapter-1-Number-Systems
- cbse-ncert-9th-class-Chapter-2-Polynomials
- cbse-ncert-9th-class-Chapter-3-Introduction-to-Euclids-Geometry
- cbse-ncert-9th-class-Chapter-4-Lines-and-Angles
- cbse-ncert-9th-class-Chapter-5-Triangles
- cbse-ncert-9th-class-Chapter-6-Coordinate-Geometry
- cbse-ncert-9th-class-Chapter-7-Herons-Formula
- cbse-ncert-9th-class-Chapter-8-Linear-Equations-in-Two-Variables
- cbse-ncert-9th-class-Chapter-9-Quadrilaterals
- cbse-ncert-9th-class-Chapter-10-Areas-of-Parallelograms-and-Triangles
- cbse-ncert-9th-class-Chapter-11-Circles
- cbse-ncert-9th-class-Chapter-12-Constructions
- cbse-ncert-9th-class-Chapter-13-Surface-Areas-and-Volumes
- cbse-ncert-9th-class-Chapter-14-Statistics
- cbse-ncert-9th-class-Chapter-15-Probability

- cbse-ncert-9th-class-Chapter-1-Number-Systems
- cbse-ncert-9th-class-Chapter-2-Polynomials
- cbse-ncert-9th-class-Chapter-3-Introduction-to-Euclids-Geometry
- cbse-ncert-9th-class-Chapter-4-Lines-and-Angles
- cbse-ncert-9th-class-Chapter-5-Triangles
- cbse-ncert-9th-class-Chapter-6-Coordinate-Geometry
- cbse-ncert-9th-class-Chapter-7-Herons-Formula
- cbse-ncert-9th-class-Chapter-8-Linear-Equations-in-Two-Variables
- cbse-ncert-9th-class-Chapter-9-Quadrilaterals
- cbse-ncert-9th-class-Chapter-10-Areas-of-Parallelograms-and-Triangles
- cbse-ncert-9th-class-Chapter-11-Circles
- cbse-ncert-9th-class-Chapter-12-Constructions
- cbse-ncert-9th-class-Chapter-13-Surface-Areas-and-Volumes
- cbse-ncert-9th-class-Chapter-14-Statistics
- cbse-ncert-9th-class-Chapter-15-Probability

- Solutions
- PDF Download
- Question Papers
- Videos

Solutions

**(i) The centre of a circle lies in ___ of the circle. (exterior/interior)****(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)****(iii) The longest chord of a circle is a ____ of the circle.****(iv) An arc is a ____ when its ends are the ends of a diameter.****(v) Segment of a circle is the region between an arc and ____ of the circle.****(vi) A circle divides the plane, on which it lies, in ____ parts.**

**Solution:**

**(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 semi-circle 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 three parts.

**(i) Line segment joining the centre to any point on the circle is a , radius of the circle.****(ii) A circle has only finite number of equal chords.****(iii) If a circle is divided into three equal arcs, each is a major arc.****(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.****(v) Sector is the region between the chord and its corresponding arc.****(vi) A circle is a plane figure.**

**Solution:**

**(i) True.**Because all points are equidistant from the centre to the circle.**(ii) False.**Because circle has infinitely may equal chords can be drawn.**(iii) False.**Because all three arcs are equal, so their is no difference between the major and minor arcs.**(iv) True.**By the definition of diameter, that diameter is twice the radius.**(v) False.**Because the sector is the region between two radii and an arc.**(vi) True.**Because circle is a part of the plane figure.

**Solution:**

Given MN and PQ are two equal chords of two congruent circles with centre at O and O’.

To prove ∠ MON = ∠ PO’Q

Proof In ∆ MON and ∆ PO’Q, we have

MO = PO’ (Radii of congruent circles)

NO = QO’ (Radii of congruent circles)

and MN = PQ (Given)

∴ By SSS criterion, we get

∆ MON = ∆ PO’Q

Hence, ∠ MON = ∠ PO’Q (By CPCT)

**Solution:**

Given MN and PQ are two chords of congruent circles such that angles subtended by .

these chords at the centres O and O’ of the circles are equal.

To prove MN = PQ

Proof In ? MON and ? PO"Q, we get

MO = PO’ (Radii of congruent circles)

NO = QO’ (Radii of congruent circles)

and ∠MON = ∠PO’Q (Given)

∴ By SAS criteria, we get

∆ MON = ∆ PO’Q

Hence, MN = PQ (By CPCT)

**Solution:**

Different pairs of circles are

**(i) Two points common**

(ii) One point is common

(iii) No point is common

(iv) No point is common

(v) One point is common

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.

Hence, a pair of circles cannot intersect each other at more than two points.

**Solution:**

**Steps of construction**

Taking three points P,Q and R on the circle.

Join PQ and QR,

Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.

Hence, O is the centre of the circle.

**Solution:**

**Given:** Two circles with centres O and O’ intersect at two points M and N so that MN is the common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.

**To prove:** OO’ is the perpendicular bisector of MN.

**Solution:**

Let O and O’ be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

**Solution:**

**Given:** RQ and MN are chords of a with centre O. MN and RQ intersect at P and MN = RQ

**To prove:**∠ OPC = ∠OPB

Solution:

Let OP be the perpendicular from O on line l. Since, the perpendicular from the centre of a circle to a chord

Now, BC is the chord of the smaller circle and OP ⊥ BC.

∴ BP = PC ..(i)

Since, AD is a chord of the larger circle and OP ⊥ AD.

∴ AP = PD .(ii)

On subtracting Eq. (i) from Eq. (ii), we get

AP - BP = PD - PC

⇒ AB = CD

Hence proved.

**Solution:**

Let O be the centre of the circle and Reshma, Salma and Mandip are represented by the points Ft, S and M, respectively.

Let RP = xm.

From Eqs. (i) and (ii), we get

Hence, the distance between Reshma and Mandip is 9.6 m.

**Solution:**

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = x

In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:

∴ ∠AOC = ∠AOB + ∠BOC = 60P + 30° = 90°

∴ Arc ABC makes 90° at the centre of the circle.

∴ ∠ADC = 12 ∠AOC

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

= 12 x 90° = 45°

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:

Let BC be chord, which is equal to the radius. Join OB and OC.

Given, BC=OB = OC

∴ ∆OBC is an equilateral triangle.

∠BOC =60°

∴ BAC = 1/2 ∠BOC

= 1/2 x 60° = 30°

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

Here, ABMC is a cyclic quadrilateral.

∴ ∠BAC + ∠BMC = 180°

(∵ In a cyclic quadrilateral the sum of opposite angles is 180°)

⇒ ∠BMC= 180° - 30° =150°

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

Solution:

∴ ∠POR = 2∠PQR = 2 x 100° = 200°

(Since, the angle subtended by the centre is double the angle subtended by circumference.)

Since, in ∆OPR, ∠POR = 360° - 200° = 160° .. (i)

Again, ∆ OPR, OP = OR (Radii of the circle)

∴ ∠OPR = ∠ORP (By property of isosceles triangle)

In ∆POR, ∠OPR + ∠ORP + ∠POR = 180° .(ii)

From Eqs. (i) and (ii), we get

∠OPR + ∠OPR + 160° = 180°

∴ 2 ∠OPR = 180° - 160° = 20°

∴ ∠OPR = 120circ/2 = 10°

In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

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

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

Solution:

Angles in the same segment are equal.

**Solution:**

**Given:** Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.

To prove: Quadrilateral NQPM is a rectangle.

Proof: ∵ ON = OP = OQ = OM (Radii of circle)

Now, ON = OP = 1/2 NP

and OM = OQ = 1/2 MQ

∴ NP = MQ

Hence, the diagonals of the quadrilateral MPQN are equal and bisect each other. So, quadrilateral NQPM is a rectangle.

**Solution:**

**Given:** Non-parallel sides PS and QR of a trapezium PQRS are equal.

∴ 2∠S + ∠Q = 360° [From Eq. (i)]

∠S+∠O = 180°

Hence, PQRS is a cyclic trape∠ium.

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 figure). Prove that ∠ ACP = ∠QCD.

Solution:

Given: 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.

To prove: ∠ACP = ∠QCD

Proof: In circle I, ∠ACP = ∠ABP (Angles in the same segment) .(i)

In circle II, ∠QCD = ∠QBD{Angles in the same segment).(ii)

∠ABP = ∠QBD (Vertically opposite angles)

From Eqs. (i) and (ii), we get ∠ACP = ∠QCD

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

Solution:

Given: Two circles are drawn with sides AC and AB of AABC as diameters . Both circles intersect each other at D.

To prove: D lies on BC.

Construction: Join AD.

Proof: Since, AC and AB are the diameters of the two circles.

∠ADB = 90° ( ∴ Angles in a semi-circle) .(i)

and ∠ADC = 90° (Angles in a semi-circle) .(ii)

On adding Eqs. (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90° = 180°

Hence, BCD is a straight line.

So, D lies on BC.

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

Solution:

Since, ∆ADC and ∆ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.

∵ Angles in the same segment are equal.

∴ ∠CBD = ∠CAD

Prove that a cyclic parallelogram is a rectangle.

Solution:

Given: PQRS is a parallelogram inscribed in a circle.

To prove: PQRS is a rectangle.

Proof: Since, PQRS is a cyclic quadrilateral.

∴ ∠P+∠R = 180°

(∵ Sum of opposite angles in a cyclic quadrilateral is 180°) .(i)

But ∠P = ∠R (∵ In a || gm opposite angles are equal) .(ii)

From Eqs. (i) and (ii), we get

∠P = ∠R = 90°

Similarly, ∠Q = ∠S = 90

∴ Each angle of PQRS is 90°.

Hence, PQRS is a rectangle.

**Solution:**

**Given:** Two circles with centres O and O’ which intersect each other at C and D.

**Solution:**

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm

∴ OM = (6 - a) cm

Join OA and OC.

Then, OA = OC = b c m

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm

In ∆OAN and ∆OCM, we get

**Solution:**

Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.

Let a be the radius of circle.

**Solution:**

Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

**Solution:**

**Given:** PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.

**To prove:** A circle drawn on PQ as diameter will pass through O.

Construction: Through O, draw MN || PS and EF || PQ.

Proof : ∵ PQ = SR ⇒ 12 PQ = 12 SR

So, PN = SM

Similarly, PE = ON

So, PN = ON = NQ

Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

**Solution:**

Since, ABCE is a cyclic quadrilateral, therefore

**Solution:**

(i) Let BD and AC be two chords of a circle bisect at P.

∴ BD divides the circle into two equal parts. So, BD is a diameter.

Similarly, AC is a diameter.

(ii) Now, BD and AC bisect each other.

So, ABCD is a parallelogram.

Also, AC = BD

∴ ABCD is a rectangle.

Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.

∴ ∠BPA = ∠BQA

(∵ Angle subtended by equal chords are equal)

⇒ BP = BQ

In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.

Solution:

(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.

Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)

and ∠BCM = ∠BAM (Angles in same segment)

But ∠BAM = ∠CAM (∵ AM is bisector of ∠A).. .(i)

∴ ∠MBC = ∠BCM

So, MB = MC (Sides opposite to equal angles are equal)

So, M must lie on the perpendicular bisector of BC

(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.

Join AM.

Since, M lies on perpendicular bisector of BC.

∴ BM = CM

∠MBC = ∠MCB

But ∠MBC = ∠MAC (Angles in same segment)

and ∠MCB = ∠BAM (Angles in same segment)

So, from Eq. (i),

∠BAM = ∠CAM

AM is the bisector of A.

Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.

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