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: Given, two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.

To Prove: ∠AOB = ∠CO’D

Proof: In ∆AOB and ∆CO’D, we have

AB = CD [Given]

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

\( \therefore \) ∆AOB ≅ ∆CO’D [By SSS congruence criteria]

\( \Rightarrow \) ∠AOB = ∠CO’D [C.P.C.T.]

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

Solution:

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 \( \cong \) ΔCOD.

\( \therefore \) By the rule of CPCT, we have

AB = CD. (Hence proved).

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

In this, there are no common points.

In this, there is only one common point.

2. Suppose you are given a circle. Give a construction to find its centre.

Solution:

Construction:

Step 1: Draw a circle with a convenient radius.

Step 2: Draw 2 chords AB and PQ of any length.

Step 3: With A as center and radii more than half the length of AB, draw two arcs on opposite sides of chord AB. With the same radius and with B as center, draw two arcs cutting the former arcs. Join the line. Now DE is the perpendicular bisector of AB. 

Step 4: With P as center and radii more than half the length of PQ, draw two arcs on opposite sides of chord PQ. With the same radius and with Q as the center, draw two arcs cutting the former arcs. Join the line. Now LM is the perpendicular bisector of PQ.

Step 5: As the center of the circle should lie both on DE and LM, it is obvious that the intersection points of DE and LM is the center of circle. Mark the intersection points as O.

Step 6: O is the required center of the circle.

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

Solution :

We have two circles with centres O and O’, intersecting at A and B.

\( \therefore \) AB is the common chord of two circles and OO’ is the line segment joining their centres.

Let OO’ and AB intersect each other at M.

\( \therefore \) To prove that OO’ is the perpendicular bisector of AB,

we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,

we have

OA = OB [Radii of the same circle]

O’A = O’B [Radii of the same circle]

OO’ = OO’ [Common]

\( \therefore \) ∆OAO’ \( \cong \) ∆OBO’ [By SSS congruence criteria]

\( \Rightarrow \) ∠1 = ∠2 , [C.P.C.T.]

Now, in ∆AOM and ∆BOM, we have

OA = OB [Radii of the same circle]

OM = OM [Common]

∠1 = ∠2 [Proved above]

\( \therefore \) ∆AOM = ∆BOM [By SAS congruence criteria]

\( \Rightarrow \) ∠3 = ∠4 [C.P.C.T.]

But ∠3 + ∠4 = 180° [Linear pair]

\( \therefore \)∠3=∠4 = 90°

\( \Rightarrow \) AM ⊥ OO’

Also, AM = BM [C.P.C.T.]

\( \Rightarrow \) M is the mid-point of AB.

Thus, OO’ is the perpendicular bisector of AB.

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 :

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

\( \because \) In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord

\( \because \) ∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

PL2 + OL2 = 2

\( \Rightarrow \) PL2 + (4 – x)2 = 52

\( \Rightarrow \) PL2 = 52 – (4 – x)2

\( \Rightarrow \) PL2 = 25 -16 – x2 + 8x

\( \Rightarrow \) PL2 = 9 – x2 + 8x …(i)

Again, in right ∆O’LP,

PL2 = PO‘2 – LO‘2

= 32 – x2 = 9 – x2 …(ii)

From (i) and (ii), we have

9 – x2 + 8x = 9 – x2

\( \Rightarrow \) 8x = 0

\( \Rightarrow \) x = 0

\( \Rightarrow \) L and O’ coincide.

\( \therefore \) PQ is a diameter of the smaller circle.

\( \Rightarrow \) PL = 3 cm

But PL = LQ

\( \therefore \) LQ = 3 cm

\( \therefore \) PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

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). here,, 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 center of the circle, draw a perpendicular to AB i.e. OM ⊥ AB

Step 2: Similarly, draw ON ⊥ CD.

Step 3: Join OE.

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)

\( \therefore \) ΔOME \( \cong \) 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: 

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.

In △ OVT and △ OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

< OVT = <OUT (Each 90o)

OT = OT (common)\( \cong ~\triangle \)

\( \therefore \)△OVT≅△ OUT   (RHS congruence rule)

\( \therefore \)<OTV = <OTU    (by CPCT)    

Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

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. 10.25).

Solution:

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

\( \therefore \) 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, 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:

Draw perpendiculars OA and OB on RS and SM respectively. Let R, S and M be the position of Reshma, Salma and Mandip respectively.

AR = AS = 3cm

OR = OS = OM = 5 m     (radii of circle)

In OAR

OA2 + AR2 = OR2            

OA2 + (3 m)2 = (5 m)2

OA2 = (25 - 9) m2 = 16 m2

OA = 4 m                           

We know that in an isosceles triangle altitude divides the base, so in △ RSM

\( \angle \) RCS will be of 90o and RC = CM     

Area of △ORS =   ½  x OA x RS

½ x RC x OS = ½ x 4 x 6

RC x 5 =24  

RC = 4.8

RM = 2RC = 2(4.8)= 9.6   

So, 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 distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution:

 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 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

\( \Rightarrow \) AD2 = AB2 -BD2

\( \Rightarrow \) AD2 = a2 -(a/2)2

\( \Rightarrow \) AD2 = 3a2/4

\( \Rightarrow \) 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.

1. Here, in figure 10.36,  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°

\( \therefore \) ∠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:

We have a circle having a chord AB equal to radius of the circle.

\( \therefore \) AO = BO = AB

\( \Rightarrow \) ∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle is 60°.

\( \Rightarrow \) ∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.

Similarly, ∠ADB = ½  [∠AOB] = ½  x 60° = 30°

Hence, the angle subtended by the chord on the major arc = 30°

3. In Fig. 10.37, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Solution:

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.

\( \therefore \) reflex ∠POR = 2∠PQR

But ∠PQR = 100°

\( \therefore \) reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

\( \Rightarrow \) ∠POR = 360° – 200°

\( \Rightarrow \) ∠POR = 160°

Since, OP = OR [Radii of the same circle]

\( \therefore \) In ∆POR, ∠OPR = ∠ORP

[Angles opposite to equal sides of a triangle are equal]

Also, ∠OPR + ∠ORP + ∠POR = 180°

[Sum of the angles of a triangle is 180°]

\( \Rightarrow \) ∠OPR + ∠ORP + 160° = 180°

\( \Rightarrow \) 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]

\( \Rightarrow \) ∠OPR = 20/2 = 10°

4.In Fig. 10.38, ∠ 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 in 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°

\( \therefore \) ∠BDC = 80°

5. In Fig. 10.39, 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 will use the following concepts to answer the question.

• The sum of angles in a triangle is 180°.
• Angles in the same segment are equal.

Consider the straight-line BD. As the line AC intersects with the line BD, the sum of two adjacent angles so formed is 180°.

Therefore, ∠BEC + ∠DEC = 180°

130° + ∠DEC = 180°

∠DEC =180° - 130° = 50°

Consider the ∆DEC, the sum of all angles will be 180º.

∠DEC + ∠EDC + ∠ECD = 180°

50° + ∠EDC + 20° = 180°

∠EDC = 180° - 70° = 110°

\( \therefore \) ∠BDC = ∠EDC = 110°

We know that angles in the same segment of a circle are equal.

\( \therefore \) ∠BAC = ∠BDC = 110°

6. 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:

Consider the chord CD,

We know that angles in the same segment are equal.

So, ∠ CBD = ∠ CAD

\( \therefore \) ∠ CAD = 70°

Now, ∠ BAD will be equal to the sum of angles BAC and CAD.

So, ∠ BAD = ∠ BAC+∠ CAD

= 30°+70°

\( \therefore \) ∠ BAD = 100°

We know that the opposite angles of a cyclic quadrilateral sums 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:

Since AC and BD are diameters.

\( \Rightarrow \) AC = BD …(i) [All diameters of a circle are equal]

Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]

Similarly, ∠ABC = 90°, ∠BCD = 90°

and ∠CDA = 90°

now, in ∆ABC and ∆BAD, we have

AC = BD [From (i)]

AB = BA [Common hypotenuse]

∠ABC = ∠BAD [Each equal to 90°]

\( \therefore \) ∆ABC ≅ ∆BAD [By RHS congruence criteria]

\( \Rightarrow \) BC = AD [C.P.C.T.]

Similarly, AB = DC

Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.

\( \therefore \) ABCD is a rectangle.

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

Solution:

We have a trapezium ABCD such that AB ॥ CD and AD = BC.

Let us draw BE ॥ AD such that ABED is a parallelogram.

\( \because \) The opposite angles and opposite sides of a parallelogram are equal.

\( \because \) ∠BAD = ∠BED …(i)

and AD = BE …(ii)

But AD = BC [Given] …(iii

\( \therefore \) From (ii) and (iii), we have BE = BC

\( \Rightarrow \) ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]

Now, ∠BED + ∠BEC = 180° [Linear pair]

\( \Rightarrow \) ∠BAD + ∠BCE = 180° [Using (i) and (iv)]

i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.

\( \therefore \) ABCD is cyclic.

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. 10.40). Prove that ∠ ACP = ∠ QCD.

Solution: 

Join chords AP and DQ

For chord AP

< PBA = < ACP     (Angles in same segment)   ... (1)  

For chord DQ

<DBQ = <QCD     (Angles in same segment)   ... (2)   

ABD and PBQ are line segments intersecting at B.

\( \therefore \)<PBA = <DBQ     (Vertically opposite angles)   ... (3)    

From equations (1), (2) and (3), we have

<ACP = <QCD

10. 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:

Consider a  \( \triangle \) ABC.

Two circles are drawn while taking AB and AC as diameter.

 Let they intersect each other at D and let D does not lie on BC.

 Join AD

    <ADB = 90o   (Angle subtend by semicircle)

    <ADC = 90o   (Angle subtend by semicircle)

    <BDC = <ADB + <ADC = 90o + 90o = 180o

 Hence BDC is straight line and our assumption was wrong.

 Thus, Point D lies on third side BC of \( \triangle \) ABC.

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

Solution:

We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC

\( \therefore \) Both the triangles are in semi-circle.

Case – I: If both the triangles are in the same semi-circle.

\( \Rightarrow \) A, B, C and D are concyclic.

Join BD.

DC is a chord.

\( \therefore \) ∠CAD and ∠CBD are formed in the same segment.

\( \Rightarrow \) ∠CAD = ∠CBD

Case – II : If both the triangles are not in the same semi-circle.

\( \Rightarrow \) A,B,C and D are concyclic. Join BD. DC is a chord.

\( \therefore \) ∠CAD and ∠CBD are formed in the same segment.

\( \therefore \) ∠CAD = ∠CBD

12. Prove that a cyclic parallelogram is a rectangle.

Solution:

We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.

\( \therefore \) Sum of its opposite angles is 180°.

\( \Rightarrow \) ∠A + ∠C = 180° …(i)

But ∠A = ∠C …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

\( \Rightarrow \) Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.

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