__PDF| Solution of Exercise: 8.1(Circle) of class 9 and 10__

## PDF file of Chapter 8.1: CLICK HERE

## Exercise 8.1 _Class – 9 and 10_

Problem – 1 : Prove that if two chords of a circle bisect each other, their point of intersection is the centre of the circle.

General enunciation: If two chords of a circle bisect each other, we have to show that their point of intersection is the centre of the circle.

Particular enunciation: Suppose two chords AC and BD of the circle ABCD bisect each other at the point O i.e., OA = OC and OD = OB. Let us prove that O is the centre of the circle.

Construction: (A,B) , (B, C), (C, D), and (D, A) are joined.

Proof: In ∆ AOD and ∆BOC we get,

OA = OC [given]

OD = OB [given]

And ∠AOD =∠BOD [vertically opposite angle]

∴∆AOD ≅ ∆BOC [by side-angle-side theorem]

Then we get, AD = BC.

Similarly, in ∆ AOB and ∆DOC we can prove that AB = CD.

Since the opposite sides of the quadrilateral ABCD are equal then it is a parallelogram.

∴AB||CD and AD||BC

Now, AB||CD and BD is their secant.

∴∠CBD = ∠ADB ————————–(1) [alternate angles]

Similarly, AD||BC and BD is their secant.

∴∠ABD = ∠CDB ————————–(2) [alternate angles]

Adding equation (1) and (2) we get,

∠CBD+∠ABD = ∠ADB+∠CDB

⟹∠ABC = ∠ADC

∴ ∠B = ∠D

Now ABCD is the quadrilateral inscribed the circle. Then the sum of the opposite angles is two right angles,

i.e., ∠ B+ ∠D = 180^{0}

⟹∠ B+ ∠ B = 180^{0} [∵∠B = ∠D]

⟹2∠ B = 180^{0}

∴ ∠B = 90^{0}

Since one of the angles of the parallelogram ABCD is 90^{0}, hence it is a rectangle and then AB = CD. Hence AC and BD are the two diameters of the circle and they bisect each other at the point O.

∴ O is the centre of the circle.

Problem – 2 : Prove that the straight line joining the middle points of two parallel chords of a circle pass through the centre and is perpendicular to the chords.

General enunciation: We have to prove that the straight line joining the middle points of two parallel chords of a circle passes through the centre and is perpendicular to the chords.

Particular enunciation: Suppose, O is the centre of the circle PQRS. Two parallel chords of it are PQ and RS. X and Y are the middle points of PQ and RS respectively. Consider we prove that, XY passing through the centre O and perpendicular to PQ and RS.

Construction: Join (O, X) and (O, Y).

Proof: Since X is the middle point of PQ, OX is the joining line of the centre and middle point of the chord PQ then

OX ⊥ PQ.

∴ ∠ OXP = ∠OXQ = 90^{0}.

Again, Y is the middle point of RS, OY is the joining line of the centre and middle point of the chord RS then

OY ⊥ RS.

∴ ∠ OYS = ∠OYR = 90^{0}.

∴ ∠ OXP = ∠OYR

But They are the alternate angles and since PQ ‖ RS then XY is their secant.

i.e., X, O and Y are collinear.

∴ XY passes through the centre O and XY ⊥ PQ and XY ⊥ RS. (Proved)

Problem-3: Two chords AB and AC of a circle subtend equal angles with the radius passing through A. Prove that

AB = AC.

General enunciation: Two chords AB and AC of a circle subtend equal angles with the radius passing through A. We have to prove that AB = AC.

Particular enunciation: Suppose O is the centre of the circle ABC. Two chords AB and AC of the circle subtend equal angles with the radius OA, i.e. ∠OAB = ∠OAC. Let us prove that AB = AC.

Construction: Let us draw two perpendiculars OM and ON to the chords AB and AC respectively.

Proof: Since O is the centre of the circle ABC and OM ⊥ AB. Then we get

AM = ½ AB.

Similarly, O is the centre of the circle ABC and ON ⊥ CD. Then we get

AN = ½ AC.

Now in **∆ **OAM and** ∆** OAN we get ∠OAM = ∠OAN [∵∠OAB = ∠OAC]

and ∠OMA = ∠ONA [∵each of them is one right angles]

and OA is common perpendicular.

∴ **∆**OAM≅**∆**OAN

Then we get AM = AN

⟹ ½ AB = ½ AC

∴ AB = AC. (Proved)

Problem – 4 : In the figure , O is the centre of a circle and Chord AB = chord AC. Prove that ∠BAO = ∠CAO.

General enunciation: In the figure, O is the centre of a circle and Chord AB = chord AC.

We have to prove that ∠BAO = ∠CAO.

Particular enunciation: Suppose O is the centre of the circle ABC. Two chords AB and AC and AB = AC. OA is the radius passing through the centre O. Let us prove that ∠BAO = ∠CAO.

Construction: Let us draw two perpendiculars OM and ON to the Chords AB and AC respectively.

Proof: In fig – i, since O is the centre of the circle and OM ⊥ AB.

Then we get, AM = ½ AB.

Similarly, O is the centre of the circle and ON ⊥ CD.

Then we get, AN = ½ AC.

But given that AB = AC.

∴ AM = AN

Now, from fig – ii, in right angle triangle **∆ **OAM and **∆**OAN

We get ∠OMA = ∠ONA [Each of them is one right angle] and OA is common hypotenuse.

∴ **∆ **OAM ≅ **∆**OAN

Then we get, ∠OMA = ∠ONA i.e., ∠BAO = ∠CAO. (Proved)

Problem – 5 : A circle passes through the vertices of a right angled triangle. Show that, the centre of the circle is the middle point of the hypotenuse.

General enunciation: A circle passes through the vertices of a right angled triangle. We have to show that, the centre of the circle is the middle point of the hypotenuse.

Particular enunciation: Let O is the centre of the circle ABC, which passes through three vertices A, B and C of the right angle triangle **∆ **ABC. Let us show that the centre of the circle O is the middle point of the hypotenuse AB.

Construction: (O, C) are joined.

Proof: In **∆ **BOC, OB = OC [radii of the same circle]

∴ ∠OCB = ∠OBC

⟹∠ACO = ∠ABC

Again, In **∆ **AOC, OA = OC [radii of the same circle]

∴ ∠OCA = ∠CAO

⟹∠BCO = ∠BAC

In **∆ **ABC, ∠ACB + ∠ABC + ∠BAC = 180^{0}

⟹ ∠ACB + ∠ACO + ∠BCO = 180^{0} [∵∠ACO = ∠ABC, BCO = ∠BAC]

⟹ ∠ACB + ∠ACB = 180^{0}

⟹ 2 ∠ACB = 180^{0}

∴ ∠ACB = 90^{0}

Since ∠ACB = 180^{0} then AB is the hypotenuse of the **∆ **ABC.

Now, AB = OA + OB

⟹ AB = OA + OA [OA and OB are radii of the same circle]

⟹ 2OA = AB

∴ OA = ½ AB

That is, O is the middle point of AB. (Showed).

Problem-6: A chord AB one of the two concentric circles intersect the other circle at points C and D. Prove that AC = BD.

General enunciation: A chord AB one of the two concentric circles intersect the other circle at points C and D. We have to prove that AC = BD.

Particular enunciation: Let O is the common centre of the circles EAB and FCD. The chords AB of EAB intersect the circle FCB at the points C and D. Let us prove that AC = BD.

Construction: Let us draw OP perpendicular to AB or CD.

Proof: Since O is the centre the circle and OP ⊥ CD. Then we get CP = PD.

Similarly, O is the centre of the circle and OP ⊥ AB. Then we get AP = PB.

But, AP = AC + CP and PB = PD + BD.

∴ AC + CP = PD + BD

⟹ AC + CP = CP + BD [∵ CP = PD]

∴ AC = BD. (Proved)

Problem -7: If two equal chords of a circle intersect each other, show that two segments of one are equal to two segments of the other.

General enunciation: If two equal chords of a circle intersect each other, show that two segments of one are equal to two segments of the other.

Particular enunciation: Suppose O is the centre of the circle MQNP. Two equal chords MN and PQ intersect each other at the point A. Let us prove that MA = PA and AQ = AN.

Construction: Let us draw two perpendiculars OC and OB to the chords MN and PQ respectively. (O, A) are joined.

Proof: Since equal chords are equidistant from the centre.

So, in right angle **∆ **COA and **∆ **BOA we get

OC = OB and OA is the common hypotenuse.

∴ **∆ **COA ≅ **∆ **BOA

∴ AC = AB —————————(i)

Now, since O is the centre of the circle and OC ⊥ MN then we get MC = ½ MN.

Similarly, O is the centre of the circle and OB ⊥ PQ then we get PB = ½ PQ.

But given that MN = PQ ——————(ii)

∴ MC = PB ————————————(iii)

Adding equations (i) and (iii) we get

AC + MC = AB + PB

⟹ AM = AP ———————————-(iv)

Again, subtracting (iv) from (ii) we get

MN – AM = PQ – AP

⟹ AN = AQ

∴ AM = AP and AN = AQ. (Proved)

Problem – 11 : Show that of the two chords of a circle, the greater chord is nearer to the centre than the shorter.

General enunciation: We have to show that of the two chords of a circle, the greater chord is nearer to the centre than the shorter.

Particular enunciation: Let, O is the centre of the circle ABCD. AB and CD are two chords where AB > CD. OE and OF are the perpendiculars from centre O to the chords AB and CD respectively. Let us show that, OE < OF.

Construction: (O, A) and (O, C) are joined.

Proof: Because O is the centre of the circle and OE ⊥ AB then we get AE = ½ AB.

Similarly, O is the centre of the circle and OF ⊥ CD then we get CF = ½ CD.

Now, in right angle triangle **∆ **OAE, OA is the hypotenuse.

∴ OA^{2} = OE^{2} + AE^{2} ………(1)

Again, in right angle triangle **∆ **OFC, OC is the hypotenuse.

∴ OC^{2} = OF^{2} + CF^{2} ………(2)

But, OA and OC are the radii of the same circle.

∴ OA = OC

So, from equation (1) and (2) we get,

OE^{2} + AE^{2} = OF^{2} + CF^{2}

⟹ AE^{2} – CF^{2} = OF^{2} – OE^{2} ……… (3)

But, according to the given question, AB > CD

⟹ ½ AB > ½ CD

⟹ AE > CF

⟹ AE^{2} >CF^{2}

⟹AE^{2} – CF^{2} > 0

⟹ OF^{2} – OE^{2} > 0 [∵AE^{2} – CF^{2} = OF^{2} – OE^{2}]

⟹ OF > OE

That is, OE < OF. (Proved)