## Extra Questions for Class 10 Maths Circles with Answers

Extra Questions for Class 10 Maths Chapter 10 Circles. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download **NCERT Solutions Class 10 Maths** to help you to revise complete syllabus and score more marks in your examinations.

### Circles Class 10 Extra Questions Very Short Answer Type

Question 1.

In fig., QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meet QR at P. If PT = 3.8 cm, find the length of QR.

Answer:

Length of tangents drawn from external point to a circle are equal.

∴ QP = PT and PR = PT

QP = 3.8 cm and PR = 3.8 cm

Now, QR = QP + PR = 3.8 + 3.8 = 7.6 cm

Question 2.

From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE 2016]

Answer:

∵ PA and PB are tangents to the given circle.

∴ ∠PAO = 90° (Radius is perpendicular to the tangent at the point of contact O.)

Now, ∠PAB = 50° (Given)

∴ ∠OAB = ∠PAO – ∠PAB

= 90° – 50° = 40°

In ∆ OAB OB = OA (Radii of the circle)

∴ ∠OAB = ∠OBA = 40°

(Angles opposite to equal sides are equal)

Now, ∠AOB + ∠OAB + ∠OBA = 180° (Angle sum property)

⇒ ∠AOB = 180° – (40° + 40°) = 100°

Question 3.

In figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA. [CBSE Outside Delhi 2016]

Answer:

Join OC, OA = OC [radii of same circle]

⇒ ∠ACO = ∠CAO = 30°

and ∠OCP = 90°

[tangent ⊥ radius at point of contact]

⇒ ∠PCA = ∠OCP – ∠AC)

= 90° – 30° = 60°

Question 4.

In fig., if AD = 15 cm, CF = 12 cm and BE = 7 cm, then find the perimeter of the triangle ABC.

Answer:

Since, the lengths of two tangents drawn from an external point to a circle are equal.

∴ AD = AE = 15 cm, BE = BF = 7 cm,

CF = CD = 12 cm

Perimeter of ∆ABC

= AB + BC + CA

= (AE + BE) + (BF + CF) + (CD + DA)

= {(15 + 7) + (7 +12) + (12 + 15)} cm

= (22 + 19 + 27) cm = 68 cm.

Question 5.

In fig., PT_{1} and PT_{2} are tangents to the circle drawn from an external point P. CD is a third tangent touching circle at Q. If PT_{2} = 12 cm and CQ = 2 cm. What is the length of PC?

Answer:

Length of tangents drawn from external point are equal.

Therefore, PT_{1} = PT2 = 12 cm

CQ = CT_{1} = 2 cm

Now, PC = PT_{1} – CT_{1} = (12 – 2) cm = 10 cm

Question 6.

Two tangents are drawn to a circle from an external point P, touching the circle at the points A and B and a third tangent intersects segment PA in C and segment PB in D and touches the circle at Q. If PA = 20 units, then find the perimeter of ∆PCD.

Answer:

Since, the length of tangents drawn from an external point of a circle are equal.

∴ PA = PB, CA = CQ and DB = DQ

Now, perimeter of ∆PCD = PC + CD + DP

= PC + (CQ + QD) + DP

= PC + (CA + DB) + PD [∵ CQ = CA and DQ = DB]

= (PC + CA) + (PD + DB)

= PA + PB

= PA + PA [∵ PB = PA]

= 2 PA = 2 × 20 = 40 units.

Hence, the perimeter of ∆PCD is 40 units.

Question 7.

PQ and PR are two tangents drawn from the point P to the circle whose centre is at O. If ∠QOR = 260°, then find ∠QPR.

Answer:

Clearly, ∠QOR of quad. PQOR is 360° – 260° = 100° and since angle between two tangents drawn from an external point to a circle are supplementary to the angle subtended by the line segments joining the point of contact at the centre.

∴ ∠QPR = 180° – 100° = 80°

Question 8.

In fig., PQ and PR are tangents drawn from P. If ∠QPR = 40°, then find ∠QSR.

Answer:

We know that, ∠QOR and ∠QPR are supplementary,

⇒ ∠QOR + ∠QPR = 180°

= 180° – 40° = 140°

Also we know that angle subtended at the centre is twice the angle subtended at the circumference.

∴ ∠QOR = 2 ∠QSR

⇒ ∠QSR = \(\frac{1}{2}\) ∠QOR = \(\frac{1}{2}\) × 140° = 70°

### Circles Class 10 Extra Questions Short Answer Type-1

Question 1.

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. [CBSE 2017]

Answer:

PA = PC + CA = PC + CQ

[∵ Length of tangents drawn from external point to circle are equal ⇒ CA = CQ]

⇒ 12 = PC + 3

⇒ PC = 9 cm

⇒ Similarly, PD= 9 cm

∴ PC + PD = 18 cm

Question 2.

In the figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD. [CBSE Delhi 2017]

Answer:

Construction: Extend AB and CD to meet at P

(i) – (ii) ⇒ PA – PB = PC – PD

⇒ AB = CD

Proved

Question 3.

In figure, a circle is inscribed in a ∆ ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. [CBSE 2016]

Answer:

AB = 12 cm

⇒ AD + BD = 12 cm …(1)

BC = 8 cm

⇒ BE + CE = 8 cm …(2)

CA = 10 cm

⇒ AF + CF = 10 cm …………. (3)

CF = CE ………… (4)

[Tangents drawn from external point to circle are equal]

Similarly

AF = AD …………. (5)

Also BD = BE ……………. (6)

Using (4) and (2), we get

BE + CF = 8 cm ………… (7)

Using (5) and (3), we get

AD + CF = 10 cm ……………. (8)

Using (6) and (1), we get

AD + BE = 12 cm …………….. (9)

Adding (7), (8) and (9), we get

BE + CF + AD + CF + AD + BE = 8cm + 10cm + 12cm

⇒ 2AD + 2BE + 2CF = 30 cm

⇒ 2(AD + BE + CF) = 30 cm

⇒ AD + BE + CF = 15 cm …….. (10)

Subtracting (7) from (10), we get

AD + BE + CF – BE – CF = 15 cm – 8 cm

⇒ AD = 7 cm

Subtracting (8) from (10), we get

AD + BE + CF – AD – CF = 15 cm – 10 cm

⇒ BE = 5 cm

Subtracting (9) from (10), we get

AD + BE + CF – AD – BE = 15 cm – 12 cm

⇒ CF = 3 cm

Thus, the lengths of AD, BE and CF are 7 cm, 5 cm and 3 cm, respectively.

Question 4.

In figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA. [A.I.2017, CBSE Outside Delhi 2016]

Answer:

As tangents drawn from external point are equal

∴ AP = AS, BP = BQ, CR = CQ, DR = DS

∴ AB + CD = AP + PB + CR + DR

⇒ AB + CD = AS + BQ + CQ + DS

= (AS + SD) + (BQ + CQ) = AD + BC

Question 5.

If PS and PT are tangents from an external point P such that PS = 10 cm and ∠SPT = 60°. Find the length of chord ST.

Answer:

As tangents from external point are equal in length.

∴ PT = PS

⇒ APST is isosceles A.

⇒ ∠PTS = ∠PST = \(\frac{\left(180^{\circ}-60^{\circ}\right)}{2}\) = 60°

⇒ ∆PST is equilateral.

∴ PS = PT = ST

∴ ST = 10 cm.

Question 6.

In the given figure, O is the centre of the circle, PT is the tangent and PAB is the secant passing through centre O. If PT = 8 cm and PA = 4 cm, then find the radius of the circle.

Answer:

Let x be the radius of circle.

⇒ OT = OA = X

PO = 4 + x

OT ⊥ PT

∆OTP is rt. ∠d at T.

∴ OP^{2} = OT^{2} + TP^{2}

⇒ (4 + x)^{2} = x^{2} + 64

⇒ 16 + x^{2} + 8x = x^{2} + 64

⇒ 8x = 48

⇒ x = 6 cm.

Question 7.

ABC is a right triangle, right angled at B. A circle is inscribed in it. The lengths of the two sides containing the right angle are 6 cm and 8 cm. Find the radius of the incircle.

Answer:

Let ‘r’ be radius of the circle. Further, let D, E, F are the points where the incircle touches the sides AB, BC, CA respectively.

Then OD = OE = OF = r cm

Also, AB = 8 cm and BC = 6 cm

Since, the tangents to a circle from an external point are equal, we have

AF = AD = (8 – r) cm

and CF = CE = (6 – r) cm

∴ AC = AF + CF = (8 – r) + (6 – r)

= 14 – 2 r

Using Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2}

⇒ (14 – 2r)^{2} = 8^{2} + 6^{2} = 100

⇒ 14 – 2r = 10

⇒ 2r = 4

⇒ r = 2 cm

Hence, the radius of incircle is 2 cm.

Question 8.

A circle touches the side BC of a ∆ABC at P, and touches AB and AC produced at Q and R respectively as shown in fig. Show that AQ = \(\frac{1}{2}\) (Perimeter of ∆ABC).

Answer:

Since, lengths of tangents drawn from an external point to a circle are equal.

∴ AQ = AR ………… (i)

BP = BQ ………… (ii)

CP = CR …………. (iii)

Perimeter of ∆ABC

= AB + BC + CA

= AB + BP + PC + AC

= AB + BQ + CR + AC

(∵ of (ii) and (iii))

= AQ + AR = 2AQ (∵ of (i))

⇒ AQ = \(\frac{1}{2}\) (Perimeter of ∆ABC)

Question 9.

In fig. FA is a tangent from an external point P to a circle with centre O. If ∠POB = 115°, then find ∠APO.

Answer:

∵ ∠POA + ∠POB = 180° (Linear pair)

∴ ∠POA = 180° – 115° = 65°

Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠PAB = 90°

Now, in ∆APO,

∠APO + ∠PAO + ∠POA = 180°

or ∠APO + 90° + 65° = 180° (Sum of angles of ∆)

or ∠APO+ 155° = 180°

or ∠APO = 180° – 155° = 25°

Question 10.

In figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60° Find the length of chord AB. [CBSE 2016]

Answer:

As the lengths of the tangents drawn from an external point to a circle are equal.

∴ PA = PB

In ∆PAB, sides PA and PB are of the same length.

Hence, ∆PAB is isosceles, with PA = PB and ∠PAB = ∠PBA = x (say).

It is given that

∠APB = 60°

In ∆PAB

∠PAB + ∠PBA + ∠APB =180°

[Angle sum property of A]

∴ x + x + 60° = 180°

⇒ 2x = 120°

⇒ x = 60°

Thus, ∠PAB = ∠PBA = ∠APB = 60°

⇒ ∆PAB is equilateral with AP = BP = AB

It is given that AP = 5 cm

∴ AB = AP = 5 cm

Thus, the length of the chord AB is 5 cm.

Question 11.

In figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. [CBSE Outside Delhi 2016]

Answer:

In ∆OTP

∠OTP = 90° [radius 1 tangent]

Let ∠OPT = ∠OPS = θ

sin θ = \(\frac{\mathrm{OT}}{\mathrm{OP}}=\frac{r}{2 r}=\frac{1}{2}\)

⇒ θ = 30°

In ∆OTP using angle sum formula for ∆

∠POT = 180° – (∠OTP + θ)

= 180° – (90° + 30°) = 60°

Similarly, ∠POS = 60°

⇒ ∠SOT = 60° + 60° = 120°

Also, in ∆OTS OT = OS = r

⇒ ∠OTQ = ∠ OSQ

⇒ ∠OTQ = ∠ OSQ = \(\left(\frac{180^{\circ}-120^{\circ}}{2}\right)\) = 30°

### Circles Class 10 Extra Questions Short Answer Type-2

Question 1.

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. [CBSE Delhi 2017]

Answer:

Given: TP, TQ are tangents to circle C(o, r) at P and Q

To prove: ∠PTQ = 2∠OPQ

Proof: Let ∠OPQ = θ

∠TPO = 90° [∵ Radius X Tangent]

⇒ ∠TPQ = ∠TPO-∠QPO = 90° – θ

Similarly ∠TQP = 90° – 0

But in ∆TPQ

∠TPQ + ∠TQP + ∠PTQ = 180° [Angle sum property]

90° – θ + 90° – θ + ∠PTQ = 180°

⇒ ∠PTQ = 2θ

⇒ = 2∠OPQ

Proved

Question 2.

Prove that the line segment joining the points of contact of two parallel tangents of a circle passes through its centre.

Answer:

Given: l and m are two parallel tangents to the circle of centre ‘O’ at points A and B respectively.

To prove: AB is a diameter of circle i.e., AOB is a straight line.

Construction: Draw a line ‘n’ through O such that, l || n || m.

Proof: Now, l|| n and AB is a transversal.

∴ ∠1 + ∠2 = 180° [Co-interior angles are supplementary] ……. (i)

But ∠1 = 90°

[Radius OA ⊥ tangent ‘l’] …….. (ii)

(i) and (ii)

⇒ 90° + ∠2 = 180°

⇒ ∠2 = 180° – 90° = 90° …….. (iii)

Similarly, m||n and AB is a transversal.

∠3 + ∠4 = 180° … (iv)

But ∠4 = 90°

[Radius OB ⊥ tangent ‘m’] ……… (v)

(iv) and (v)

⇒ ∠3 + 90° = 180°

⇒ ∠3 = 180° – 90° = 90° …….. (vi)

Now ∠AOB = ∠2 + ∠3 = 90° + 90° = 180°

[Using (iii) and (vi)]

∴ AOB is a straight line.

Hence, AB is diameter of circle.

So, line segment joining the point of contact of two parallel tangents of a circle passes through its centre.

Question 3.

If from an external point P of a circle with centre O two tangents PQ and PR are drawn such that ∠QPR = 120°. Prove that 2PQ = PO.

Answer:

In ∆POQ and ∆POR.

∠1 = ∠2 = 90°

[90° = Angle between tangent and radius]

PO = OP (Common)

PQ = PR (Tangents drawn from external point are equal)

∴ ∆POQ ≅ ∆POR (By RHS)

⇒ ∠3 = ∠4 = \(\frac{1}{2}\) ∠QPR

Question 4.

ABC is a right-angled triangle, right angled at B and with BC = 6 cm and AB = 8 cm. A circle with centre O and radius x has been described in AABC. Find the values of x.

Answer:

We know that radius is perpendicular to tangent.

PO ⊥ BC and OQ ⊥ AB

∴ OPBQ is a rectangle.

But OP = OQ = x

OPBQ is a square

Now, BP = x

∴ PC = (6 – x) cm

∴ OP = QB = x

∴ AQ (8 – x) cm

and PC = CR = (6 – x) cm

Now, AC = \(\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}\)

(By Pythagoras Theorem)

= \(\sqrt{8^2+6^2}\) = \(\sqrt{64^2+36^2}\) = √100

= 10 cm

AC = AR + CR

10 = (8 – x) + (6 – x)

10 = 14 – 2x

2x = 4cm

x = 2 cm

Question 5.

In the given figure ‘O’ is the centre of the circle. Determine ∠AQB and ∠AMB, if PA and PB are tangents and, ∠APB = 75°

Answer:

Given: ∠APB = 75°, PA and PB are tangents.

To find: ∠AQB and ∠AMB

Proof: ∠AOB = 180° – 75° = 105° [∵ ∠P + ∠AOB = 180°]

∴ Reflex ∠AOB = 360° -105° = 255°

∴ ∠AQB = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) x 255°

= 127\(\frac{1^{\circ}}{2}\)

Question 6.

In the figure, quadrilateral ABCD circumscribes the circle. Find the length of the side CD.

Answer:

AE = AH

[Length of tangents from external points]

x = 4 – x

⇒ 2x = 4

⇒ x = 2

DH = (5 – 2) = 3 cm

DH = DG = 3 cm

CF = CG

⇒ 2y – 3 = y

⇒ y = 3

∴ DC = DG + GC = 3 + 3 = 6 cm.

### Circles Class 10 Extra Questions Long Answer Type 1

Question 1.

Prove that lengths of tangents drawn from an external point to a circle are equal. [CBSE 2017, 18]

Answer:

Given: A circle C(o, r), PA & PB are tangents drawn from external point P to the circle.

To prove: PA = PB

Construction: Join OA, OB & OP

Proof: Since radius ⊥ tangent at the point of contact.

⇒ ∠PAO = ∠PBO ……………. (i)

In ∆OAP & ∆OBP

OA = OB

OP = OP

∠PAO = ∠PBO

⇒ ∆OAP ≅ ∆OBP

⇒ PA = PB [Hence Proved] c.p.c.t.

Question 2.

Prove that the lengths of tangents drawn from an external point to a circle are equal. [CBSE 2016]

Using the above, do the following:

In figure TP and TQ are tangents from T to the circle with centre O and R is any point on the circle. If AB is tangent to the circle at R, prove that TA + AR = TB + BR.

Answer:

Refer Theorem II in Synopsis for first part.

Next,

TP = TQ [∵ Theorem II]

⇒ TA + AP = TB + BQ ………. (i)

But AP = AR ………………. (ii) [AP, AR are tangents drawn from external point A]

Similarly, BQ = BR ……………. (iii)

Putting for AP and BQ in (i) from (ii) and (iii) we get

TA + AR = TB + BR (Proved)

Question 3.

The centre of a circle of radius 13 cm is the point C (3, 6) and P (7, 9). is a point inside the circle. APB is a chord of the circle such that AP = PB. Calculate the length of AB.

Answer:

Given that AC = 13 cm.

Also, C (3, 6) and P (7,9)

∴ CP = \(\sqrt{(7-3)^2+(9-6)^2}\) (Distance formula)

= \(\sqrt{16+9}\) = √25 = 5 cm

Also, CP ⊥ APB

∴ In right angled ∆CPA,

AC^{2} = AP^{2} + CP^{2} [By Pythagoras Theorem]

or (13)^{2} = AP^{2} + (5)^{2}

or AP^{2} = 169 – 25 = 144

or AP^{2} = (12)^{2}

or AP = 12

But AP = PB …………….. (Given)

⇒ AB = 2AP = 2 (12) cm = 24 cm.

Question 4.

AB and CD are two parallels of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distances between them is 17 cm, find the radius of the circle.

Answer:

Let radius of the circle be r cm.

Draw OE ⊥ CD and OF ⊥ AB.

Join OB and OD.

Also, OF = xcm

OE = (17 – x) cm

AB = 10 cm (Given)

∴ FB = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 10 = 5 cm

Similarly, ED = \(\frac{1}{2}\) CD = \(\frac{1}{2}\) (24) = 12 cm

In right angled ∆OFB,

OB^{2} = OF^{2} + FB^{2}

r^{2} = x^{2} + (5)^{2}

r^{2} = x^{2} + 25 ……………. (1)

Also, in right angled AOED,

OD^{2} = OE^{2} + ED^{2}

r^{2} = (17 – x)^{2} + (12)^{2}

r^{2} = 289 + x^{2} – 34x + 144

r^{2} = x^{2} – 34x + 433 ……………… (2)

From (1) and (2), we get

x^{2} + 25 = x^{2} – 34x + 433

34x = 408

x = \(\frac{408}{34}\) = 12

Putting the value of x in (1), we get

r^{2} = (12)^{2} + 25 = 144 + 25 = (13)^{2}

r = 13 cm

Hence, radius of the circle is 13 cm.

Question 5.

Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.

Answer:

Two circles having centre O and O’ and OA = 10 cm; O’A = 8 cm respectively.

Also, AB = 12 cm be the length of common chord.

∴ AM = \(\frac{1}{2}\) = \(\frac{1}{2}\) (12) = 6 cm

In right angled ∆OMA,

OA^{2} = OM^{2} + AM^{2}

(10)^{2} = OM^{2} + (6)^{2}

or OM^{2} = 100 – 36

or OM^{2} = 64 = (8)^{2}

or OM = 8 cm

Now, in right ∆ O’MA,

O’A^{2} = O’M^{2} + AM^{2}

(8)^{2} = O’M^{2} + (6)^{2}

or O’M^{2} = 64 – 36 = 28

or O’M = √28 = 5.29 cm.

∴ Required, distance between the centres

= OO’ = OM + MO’

= (8 + 5.29) cm

= 13.29 cm.

Question 6.

In the given figure, AB is the chord of a circle with centre O. AB is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ∠ACD = y° and ∠AOD = x°, prove that x° = 3y°.

Answer:

Given that,

BC = OB

⇒ ∠OCB = ∠BOC = y°.

In ∆OBC,

Exterior ∠OBA = ∠BOC +∠OCB

= y° + y° = 2y°

Now, OA = OB (Radii of same circle)

⇒ ∠OAB = ∠OBA = 2y°

In ∆AOC,

Exterior ∠AOD = ∠OAC + ∠OCA

= 2y° + y° = 3y°

But, ∠AOD = x° (Given that)

∴ x° = 3y°

Question 7.

In figure, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\). [CBSE Outside Delhi 2016]

Answer:

Given: Two circles C(O’, r) and C(O, r).

AX is diameter of C(O’, r) and AC is tangent to C (O, r).O’D ⊥ AC

To find: \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\)

In ∆AO’D and ∆AOC

∠A = ∠A [Common angle]

∠ADO’ = ∠ACO [90° each]

⇒ ∆AO’D ~ ∆AOC

⇒ \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}\) ……………. (1)

But AO’ = r

AO = AO’ + O’X + XO

= r + r + r = 3r

(1) ⇒ \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{20}{3}=\frac{r}{3 r}=\frac{1}{3}\)

∴ \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{1}{3}\)

Question 8.

In figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. [CBSE 2016]

Answer:

From the given figure, we have TP = TQ

[∴ Tangents, drawn form an external point to a circle, have equal length.]

and ∠TQO = ∠TPO = 90°

[∵ Tangent to a circle is perpendicular to the radius through the point of contact]

In ∆ TOQ,

QT^{2} + OQ^{2} = OT^{2}

⇒ QT^{2} = 13^{2} – 5^{2} = 144

⇒ QT = 12 cm

Now, OT – OE = ET

= 13 – 5 = 8 cm

Let QB = x cm

∴ QB = EB = x

[∵ Tangents, drawn from an external point to a circle, have equal length.]

Also, ∠OEB = 90°

[Tangent to a circle is perpendicular to the radius through the point of contact.]

In ∆TEB EB^{2} + ET^{2} = TB^{2}

⇒ x^{2} + 8^{2} = (12 – x)^{2}

⇒ x^{2} + 64 = 144 + x^{2} – 24x

⇒ 24x = 80

⇒ x = \(\frac{80}{24}=\frac{10}{3}\)

AB = 2x = cm

∴ Thus, the length of AB is \(\frac{20}{3}\) cm.

### Circles Class 10 Extra Questions HOTS

Question 1.

Let A be one point of intersection of two intersecting circles with centres O and Q. The tangents at A to the two circles meet the circles again at B and C respectively. Let the point P be located so that AOPQ is a parallelogram. Prove that P is the circumcentre of the triangle ABC.

Answer:

AQ is the radius and AB is a tangent to the circle with centre Q.

⇒ AQ ⊥ AB

[Radius is perpendicular to tangent through the point of contact]

Again OPQA is a || gm.

⇒ AQ || OP

Now AQ || OP and AQ ⊥ AB

⇒ OP⊥ AB

Let OP intersect AB at M

∴ OM ⊥ AB

∴ AM = BM

[∵ Perpendicular from the centre of a circle to a chord bisects the chord.]

∴ OM and hence OP is perpendicular bisector of AB.

Similarly, we can show that PQ is perpendicular bisector of AC.

Now in ∆ABC,

OP is the perpendicular bisector of side AB.

∴ PA = PB – [Any point on the perpendicular bisector of line segment is equidistant from the end points.]

Similarly, PA = PC

PA = PB = PC

⇒ P is equidistant from three vertices of ∆ABC.

⇒ The circle with P as centre and its distance from any vertex as radius passes through the three vertices of ∆ ABC and the point P is the circumcentre of the ∆ABC.

Question 2.

In the adj oining figure from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove that

(i) PA.PB = PN^{2} – AN^{2}

(ii) PN^{2} – AN^{2} = OP^{2} – OT^{2}

(iii) PA.PB = PT^{2}

Answer:

Observe that PA = PN – AN and PB = PN + BN

(i) ∴ PA. PB = (PN – AN) (PN + BN)

= (PN – AN) (PN + AN)

[AN = BN as perpendicular from centre bisects the chord]

= PN^{2} – AN^{2} ……………… (i)

(ii) In right ∆ONP,

OP^{2} = ON^{2} + NP^{2} ⇒ NP^{2}

= OP^{2} – ON^{2} ………….. (a)

and in right ^{2}ONA,

ON^{2} + AN^{2} = OA^{2} ………….. (b)

∴ PN^{2} – AN^{2} = (OP^{2} – ON^{2}) – AN^{2} [using (a)]

= OP^{2} – (ON^{2} + AN^{2})

= OP^{2} – OA^{2} [using (b)] …………….. (ii)

= OP^{2} – OT^{2} [∵ OA = OT]

(iii) From (i) and (ii)

⇒ PA. PB = OP^{2} – OT^{2} = PT^{2}

[in right ∠d ∆, OP^{2} = OT^{2} + PT^{2}].

Question 3.

In the adjoining fig., AB is a line segment ’ and M is its mid-point. Semicircles are drawn with AM, MB and AB as diameters on the same side of line AB. A circle C(O, r) is drawn so that it touches all the three semicircles.

Prove that r = \(\frac{1}{6}\) AB.

Answer:

Let us mark ‘V and ‘N’ as mid-points of AM and MB. Join OL and ON.

Let AB = x

Note that O, P, L, are collinear and the points O, Q, N are collinear.

In rt. ∠d AOML

OL^{2} = OM^{2} + LM^{2}

Question 4.

From point P outside the circle, with centre O, tangents PA and PB are drawn in the adjoining fig. If OP is equal to the diameter of the circle, prove that PAB is an equilateral triangle.

Answer:

Let us join OA, OB, AQ

Now, OP = 2OQ [∵ Diameter = 2 Radius]

⇒ OQ = PQ …….. (i)

or Q is mid-point of OP.

Also, OA ⊥ PA [Radius ⊥ tangent through the point of contact]

∆OAP is a right angled triangle.

Q is mid-point of hypotenuse OP.

⇒ QA = OQ = PQ

[As mid-point of hypotenuse of a right triangle is equidistant from three vertices] …… (ii)

But OQ = OA [Radii of same circle] ……. (iii)

(ii) and (iii)

⇒ OA = OQ = QA

⇒ ∆OAQ is an equilateral triangle.

⇒ ∠AOQ = 60°

But in ∆OAP,

∠APO + ∠OAP + ∠AOP = 180°

⇒∠APO + 90° + 60° = 180° [∴ ∠AOQ = ∠AOP = 60°]

⇒ ∠APO = 180° – 150° = 30°

Similarly we show that,

∠BPO = 30°

⇒ ∠APB = ∠APO + ∠BPO

= 30°+ 30° = 60° …(iv)

Aso in ∆APB,

PA = PB

[Tangents from external point P]

∠PAB = ∠PBA ………….. (v)

In ∆APB, using angle sum property of triangle.

∠PAB + ∠PBA + ∠APB = 180°

⇒ 2∠PAB + 60° = 180°

⇒ 2∠PAB = 180° – 60° = 120°

⇒ ∠PAB = 60° ………….. (vi)

(iv), (v) and (vi)

⇒ ∠APB = ∠PAB = ∠PBA = 60° each.

⇒ ∆PAB is an equilateral ∆.

Question 5.

Two circles with radii a and b touch each other externally. Let c be the radius of a circle which touches these two circles as well as a common tangent to the two circles, prove that:

\(\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\)

Answer:

Let us draw a line through centre C of smaller circle parallel to MN let it meet AM and BN at P and Q respectively. Join AC and BC. clearly ∠APC = 90° and ∠BQC = 90°

AP = AM – PM = a – c, AC = a + c

and BQ = BN – QN = b – c, BC = b + c

In right angled ∆APC,

PC = \(\sqrt{\mathrm{BC}^2-\mathrm{BQ}^2}\) = \(\sqrt{(b+c)^2-(b-c)^2}\)

= √4bc = 2√ac

Similarly, in right ∆BQC,

QC ⇒ \(\sqrt{\mathrm{BC}^2-\mathrm{BQ}^2}\) = \(\sqrt{(b+c)^2-(b-c)^2}\)

= √4bc = 2√bc

Also, draw BT || MN meeting AM at T.

Now, AB = a + b and AT = AM – MT = a – b

In right ∆ABT,

BT = \(\sqrt{\mathrm{AB}^2-\mathrm{AT}^2}\) = \(\sqrt{(a+b)^2-(a-b)^2}\)

= √4ab = 2√ab

But BT = MN

⇒ MN = 2√ab

But MN = MR + RN = PC + QC

⇒ 2√ab = 2√ac + 2√bc

Dividing throughout by 2√abc, we get

\(\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{a}}\)

Question 6.

In the given figure, AOC is a diameter of the circle. If AB = 7 cm, BC = 6 cm and CD = 2 cm, find the perimeter of cyclic quadrilateral ABCD.

Answer:

AOC is a diameter of the circle.

∴ ∠ABC = 90° (Angle in semicircle)

So, in rt. angled ∆ ABC,

AC^{2} = AB^{2} + BC^{2}

or AC^{2} = (7)^{2} + (6)^{2} = 49 + 36

or AC^{2} = 85 ……….. (1)

Similarly, ∠ADC = 90° (Angle in semicircle)

So, in right angled ∆ADC,

AC^{2} = AD^{2} + DC^{2}

or 85 = AD^{2} + (2)^{2}

or AD^{2} = 85 – 4 = 81

or AD^{2} = (9)^{2}

or AD = 9 cm

Now, perimeter of cyclic quadrilateral ABCD

= AB + BC + CD + DA

= (7 + 6 + 2 + 9) cm

= 24 cm.

Question 7.

In ∆ABC, AB = 8 cm, BC = 6 cm, CA = 4 cm. With the vertices of triangle as centre, three circles are described, each touching the other two externally. Find the radii of each circle.

Answer:

Let x cm, y cm, z cm be the radii of circle I, II and III respectively as shown in the figure.

∴ x + y = 6 cm …(1)

y + z = 4 cm …(2)

z + x = 8 cm …(3)

Adding (1), (2), (3), we get

2 (x + y + z) = 18

x + y + z = 9 …(4)

(4) – (1) gives, z = 3

(4) – (2) gives, x = 5

(4) – (3) gives, y = 1

∴ Radii of circles are 5 cm, 1 cm and 3 cm.

Multiple Choice Questions

Choose the correct option for each of the following:

Question 1.

In fig., if the semiperimeter of ∆ ABC = 23 cm, then AF + BD + CE is:

(a) 46 cm

(b) 11.5 cm

(c) 23 cm

(d) 34.5 cm

Answer:

(b) 11.5 cm

Question 2.

In fig., AP = 2 cm, BQ = 3 cm and RC = 4 cm, then the perimeter of ∆ ABC (in cm) is:

(a) 16

(b) 18

(c) 20

(d) 21

Answer:

(b) 18

Question 3.

In fig., two circles with centres A and B touch each other externally at point R.

The length of PQ (in cm) is:

(a) 27

(b) 18

(c) 24

(d) 20

Answer:

(a) 27

Question 4.

In fig., two concentric circles with centre O are shown. AB and APQ are tangents to the inner circle from point A lying on the outer circle. If AB = 7.5 cm, then AQ is equal to:

(a) 18 cm

(b) 15 cm

(c) 12 cm

(d) 10 cm

Answer:

(b) 15 cm

Question 5.

Quadrilateral ABCD circumscribes a circle as shown in fig. The side of quadrilateral which is equal to AP + BR is :

(a) AD

(b) BC

(c) AB

(d) BQ

Answer:

(c) AB

Question 6.

A tangent PT is drawn from an external point P to a circle of radius 3√2 cm such that distance of the point P from O is 6 cm as shown in fig. The value of ∠TPO is:

(a) 30°

(b) 45°

(c) 60°

(d) 75°

Answer:

(b) 45°

Question 7.

In fig., from an external point T, TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is

(a) 60°

(b) 70°

(c) 80°

(d) 90°

Answer:

(b) 70°

Question 8.

In fig., measure of ∠QSR is

(a) 60°

(b) 100°

(c) 110°

(d) 120°

Answer:

(d) 120°

Question 9.

In fig., AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to:

(a) 65°

(b) 60°

(c) 50°

(d) 40°

Answer:

(c) 50°

Question 10.

In fig., if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to:

(a) 100°

(b) 80°

(c) 90°

(d) 75°

Answer:

(a) 100°

Question 11.

In fig., if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to:

(a) 25°

(b) 30°

(c) 40°

(d) 50°

Answer:

(a) 25°

Question 12.

PQ and PT are tangents to a circle with centre O and radius 5 cm. If OP = 13 cm, then perimeter of quadrilateral PQOT is:

(a) 24 cm

(b) 34 cm

(c) 17 cm

(d) 20 cm

Answer:

(b) 34 cm

Question 13.

In fig., if OA = 5 cm and OM = 3 cm, the length of the chord AB (in cm) is:

(a) 8

(b) 10

(c) 6

(d) 4

Answer:

(a) 8

Question 14.

In fig., AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to:

(a) 4 cm

(b) 2 cm

(c) 2√3 cm

(d) 4√3 cm

Answer:

(c) 2√3 cm

Question 15.

In fig., a quadrilateral ABCD is drawn to circumscribe a circle. Then

(a) AD + BC = AB + CD

(b) AB + BC = AD + CD

(c) BC + CD = AD + AB

(d) AB + BC + CD + AD = AC + BD

Answer:

(a) AD + BC = AB + CD

Fill in the Blanks

Question 1.

A line can intersect a circle at the most at _____________ points.

Answer:

two

Question 2.

A line intersecting the circle at two points is called a _____________ .

Answer:

secant

Question 3.

A line intersecting the circle exactly at one point is called a _____________ .

Answer:

tangent

Question 4.

Portion of secant intercepted by a circle is called _____________ of the circle.

Answer:

chord

Question 5.

The common point of tangent to a circle and the circle is called the _____________ .

Answer:

point of contact

Question 6.

Every point of the either in the interior or on the circle _____________ .

Answer:

chord

Question 7.

Every point of a to the circle except the point of _____________ lie outside the circle.

Answer:

tangent, contact

Question 8.

The tangent at any point of a circle is _____________ to the radius through the point of contact.

Answer:

perpendicular

Question 9.

A circle can have _____________ parallel tangents at the most whereas it can have parallel secants.

Answer:

two, infinite

Question 10.

We can draw no tangent from a point lying _____________ the circle.

Answer:

in the interior of