## Important Questions for Class 10 Maths Chapter 10 Circles

### Circles Class 10 Important Questions Very Short Answer (1 Mark)

**Circles Class 10 Questions With Solutions Question 1.**

In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If âˆ AOB = 100Â°, then calculate âˆ BAT. (2011D)

Solution:

âˆ 1 = âˆ 2

âˆ 1 + âˆ 2 + 100Â° = 180Â°

âˆ 1 + âˆ 1 = 80Â°

â‡’ 2âˆ 1 = 80Â°

â‡’ âˆ 1 = 40Â°

âˆ 1 + âˆ BAT = 90Â°

âˆ BAT = 90Â° – 40Â° = 50Â°

**Circles Class 10 Important Questions Question 2.**

In the given figure, PA and PB are tangents to the circle with centre O. If âˆ APB = 60Â°, then calculate âˆ OAB, (2011D)

Solution:

âˆ 1 = âˆ 2

âˆ 1 + âˆ 2 + âˆ APB = 180Â°

âˆ 1 + âˆ 1 + 60Â° = 180Â°

2âˆ 1 = 180Â° – 60Â° = 120Â°

âˆ 1 = \(\frac{120^{\circ}}{2}\) = 60Â°

âˆ 1 + âˆ OAB = 90Â°

60Â° +âˆ OAB = 90Â°

âˆ OAB = 90Â° – 60Â° = 30Â°

**Class 10 Circles Important Questions Question 3.**

In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If âˆ POQ = 70Â°, then calculate âˆ TPQuestion (2011OD)

Solution:

âˆ 1 = âˆ 2

âˆ 1 + âˆ 2 + 70Â° = 180Â°

âˆ 1 + âˆ 1 = 180Â° – 70Â°

2âˆ 1 = 110Â° â‡’ âˆ 1 = 55Â°

âˆ 1 + âˆ TPQ = 90Â°

55Â° + âˆ TPQ = 90Â°

â‡’ âˆ TPQ = 90Â° – 55Â° = 35Â°

**Circles Class 10 Important Questions With Solutions Question 4.**

A chord of a circle of radius 10 cm subtends a right angle at its centre. Calculate the length of the chord (in cm). (2014OD)

Solution:

AB^{2 }= OA^{2} + OB^{2} …[Pythagoras’ theorem

AB^{2} = 10^{2} + 10^{2}

AB^{2} = 2(10)^{2}

AB = \(10 \sqrt{2}\) cm

**Circles Important Questions Class 10 Question 5.**

In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and âˆ CAB = 30Â°. Find âˆ PCA. (2016OD)

Solution:

âˆ ACB = 90Â° …[Angle in the semi-circle

In âˆ†ABC,

âˆ CAB + âˆ ACB + âˆ CBA = 180Â°

30 + 90Â° + âˆ CBA = 180Â°

âˆ CBA = 180Â° – 30Â° – 90Â° = 60Â°

âˆ PCA = âˆ CBA …[Angle in the alternate segment

âˆ´ âˆ PCA = 60Â°

**Circle Questions Class 10 Question 6.**

In the given figure, AB and AC are tangents to the circle with centre o such that âˆ BAC = 40Â°. Then calculate âˆ BOC. (2011OD)

Solution:

AB and AC are tangents

âˆ´ âˆ ABO = âˆ ACO = 90Â°

In ABOC,

âˆ ABO + âˆ ACO + âˆ BAC + âˆ BOC = 360Â°

90Â° + 90Â° + 40Â° + âˆ BOC = 360Â°

âˆ BOC = 360 – 220Â° = 140Â°

**Circle Class 10 Important Question Question 7.**

In the given figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K&M respectively. If EK = 9 cm, calculate the perimeter of AEDF (in cm). (2012D)

Solution:

Perimeter of âˆ†EDF

= 2(EK) = 2(9) = 18 cm

**Circles Class 10 Questions Question 8.**

In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP (in cm). (2012OD)

Solution:

2AP = Perimeter of âˆ†

2AP = 5 + 6 + 4 = 15 cm

AP = \(\frac{15}{2}\) = 7.5 cm

**Circles Questions Class 10 Question 9.**

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA âŠ¥ PB, then find the length of each tangent. (2013D)

Solution:

Construction: Join AC and BC.

Proof: âˆ 1 = âˆ 2 = 90Â° ….[Tangent is I to the radius (through the point of contact

âˆ´ APBC is a square.

Length of each tangent

= AP = PB = 4 cm

= AC = radius = 4 cm

**Circles Class 10 Important Questions Pdf Question 10.**

In the given figure, PQ and PR are two tangents to a circle with centre O. If âˆ QPR = 46Â°, then calculate âˆ QOR. (2014D)

Solution:

âˆ OQP = 900

âˆ ORP = 90Â°

âˆ OQP + âˆ QPR + âˆ ORP + âˆ QOR = 360Â° …[Angle sum property of a quad.

90Â° + 46Â° + 90Â° + âˆ QOR = 360Â°

âˆ QOR = 360Â° – 90Â° – 46Â° – 90Â° = 134Â°

**Important Questions Of Circles Class 10 Question 11.**

In the given figure, PA and PB are tangents to the circle with centre O such that âˆ APB = 50Â°. Write the measure of âˆ OAB. (2015D)

Solution:

PA = PB …[âˆµ Tangents drawn from external point are equal

âˆ OAP = âˆ OBP = 90Â°

âˆ OAB = âˆ OBA … [Angles opposite equal sides

âˆ OAP + âˆ AOB + âˆ OBP + âˆ APB = 360Â° … [Quadratic rule

90Â° + âˆ AOB + 90Â° + 50Â° = 360Â°

âˆ AOB = 360Â° â€“ 230Â°

= 130Â°

âˆ AOB + âˆ OAB + âˆ OBA = 180Â° … [âˆ† rule

130Â° + 2âˆ OAB = 180Â° … [From (i)

2âˆ OAB = 50Â°

â‡’ âˆ OAB = 25Â°

**Class 10th Circles Extra Questions Question 12.**

From an external point P, tangents PA and PB are drawn to a circle with centre 0. If âˆ PAB = 50Â°, then find âˆ AOB. (2016D)

Solution:

PA = PB …[âˆµ Tangents drawn from external point are equal

âˆ PBA = âˆ PAB = 50Â° …[Angles equal to opposite sides

In âˆ†ABP, âˆ PBA + âˆ PAB + âˆ APB = 180Â° …[Angle-sum-property of a âˆ†

50Â° + 50Â° + âˆ APB = 180Â°

âˆ APB = 180Â° – 50Â° – 50Â° = 80Â°

In cyclic quadrilateral OAPB

âˆ AOB + âˆ APB = 180Â° ……[Sum of opposite angles of a cyclic (quadrilateral is 180Â°

âˆ AOB + 80o = 180Â°

âˆ AOB = 180Â° – 80Â° = 100Â°

**Circles Important Questions Question 13.**

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If âˆ QPT = 60Â°, find âˆ PRQ. (2015OD)

Solution:

PQ is the chord of the circle and PT is tangent.

âˆ´ âˆ OPT = 90Â° …[Tangent is I to the radius through the point of contact

Now âˆ QPT = 60Â° … [Given

âˆ OPQ = âˆ OPT – âˆ QPT

â‡’ âˆ OPQ = 90Â° – 60Â° = 30Â°

In âˆ†OPQ, OP = OQ

âˆ OQP = âˆ OPQ = 30Â° … [In a âˆ†, equal sides have equal âˆ s opp. them

Now, âˆ OQP + âˆ OPQ + âˆ POQ = 180Â°

âˆ´ âˆ POQ = 120Â° …[âˆ POQ = 180o â€“ (30Â° + 30Â°)

â‡’ Reflex âˆ POQ = 360Â° – 120Â° = 240Â° …[We know that the angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle

âˆ´ Reflex âˆ POQ = 2âˆ PRO

â‡’ 240Â° = 2âˆ PRQ

â‡’ âˆ PRQ = \(\frac{240^{\circ}}{2}\) = 120Â°

**Class 10 Circles Questions Question 14.**

In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). (2012D)

Solution:

AP = AR = 4 cm

RC = 11 – 4 = 7 cm

RC = QC = 7 cm

BQ = BP = 3 cm

BC = BQ + QC

= 3 + 7 = 10 cm

**Circle Class 10 Questions With Solutions Question 15.**

In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. Calculate the radius of the circle inscribed in the triangle (in cm). (2014OD)

Solution:

AC^{2} = AB^{2} + BC^{2} …[Pythagoras’ theorem

= (5)^{2} + (12)^{2}

AC^{2} = 25 + 144

AC = \(\sqrt{169}\) = 13 cm

Area of âˆ†ABC = Area of âˆ†AOB + ar. of âˆ†BOC + ar. of âˆ†AOC

60 = r(AB + BC + AC)

60 = r(5 + 12 + 13)

60 = 30r â‡’ r = 2 cm

Question 16.

Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. (2011OD)

Solution:

Radius = R

AB = a + a = 2a

âˆ´ Perimeter = 4(AB)

= 4(2a)

= 8a cm

Question 17.

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius DA of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD âŠ¥ CD, then calculate the length of CD. (2013OD)

Solution:

Const. Join OR

Proof. âˆ 1 = âˆ 2 = 90Â° … [Tangent is âŠ¥ to the radius through the point of contact

âˆ 3 = 90Â° …[Given

âˆ´ ORDS is a square.

DR = OS = 10 cm …(i)

BP = BQ = 27 cm …[Tangents drawn from an external point

âˆ´ CQ = 38 – 27 = 11 cm

RC = CO = 11 cm …[Tangents drawn from an external point

DC = DR + RC = 10 + 11 = 21 cm …[From (i) & (ii)

### Circles Class 10 Important Questions Short Answer-I (2 Marks)

Question 18.

Prove that the tangents drawn at the ends of a diameter of a circle are parallel. (2012OD)

Solution:

Proof: âˆ 1 = 90Â° …(i)

âˆ 2 = 90Â° …(ii)

âˆ 1 = âˆ 2 … [From (i) & (ii)

But these are alternate interior angles

âˆ´PQ || RS

Question 19.

In the figure, AB is the diameter of a circle with centre O and AT is a tangent. If âˆ AOQ = 58Â°, find âˆ ATQ.Â (2015D)

Solution:

âˆ ABQ = \(\frac{1}{2}\) âˆ AOQ = \(\frac{58^{\circ}}{2}\) = 29Â°

âˆ BAT = 90Â° ….[Tangent is âŠ¥ to the radius through the point of contact

âˆ ATQ = 180Â° – (âˆ ABQ + âˆ BAT)

= 180 – (29 + 90) = 180Â° – 119Â° = 61Â°

Question 20.

Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r. (2011D)

Solution:

Given: OC = 7 cm, AB = 48 cm

To find: r = ?

âˆ OCA = 90Â° ..[Tangent is âŠ¥ to the radius through the point of contact

âˆ´ OC âŠ¥ AB

AC = \(\frac{1}{2}\) (AB) … [âŠ¥ from the centre bisects the chord

â‡’ AC = \(\frac{1}{2}\) (48) = 24 cm

In rt. âˆ†OCA, OA^{2} = OC^{2} + AC^{2} … [Pythagoras’ theorem

r^{2} = (7)^{2} + (24)^{2}

= 49 + 576 = 625

âˆ´ r= \(\sqrt{625}\) = 25 cm

Question 21.

In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. (2012D)

Solution:

Const.: Join OC

Proof: AB is a tangent to smaller circle and OC is a radius.

âˆ´ âˆ OCB = 90Â° … above theorem

In the larger circle, AB is a chord and OC âŠ¥ AB.

âˆ´ AC = CB … [âŠ¥ from the centre bisects the chord

Question 22.

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD. (2011OD)

Solution:

AB + CD = AD + BC

6 + 8 = AD + 9

14 – 9 = AD â‡’ AD = 5 cm

Question 23.

Prove that the parallelogram circumscribing a circle is a rhombus. (2012D, 2013D)

Solution:

Given. ABCD is a à¥¥^{gm}.

To prove. ABCD is a rhombus.

Proof. In à¥¥^{gm}, opposite sides are equal

AB = CD

and AD = BC ..(i)

AP = AS …[Tangents drawn from an external point are equal in length

PB = BQ

CR = CO

DR = DS

By adding these tangents,

(AP + PB) + (CR + DR) = AS + BQ + CQ + DS

AB + CD = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AB + AB = BC + BC … [From (i)

2AB = 2 BC

AB = BC …(ii)

From (i) and (ii), AB = BC = CD = DA

âˆ´ à¥¥^{gm} ABCD is a rhombus.

Question 24.

In figure, a quadrilateral ABCD is drawn to circum- DA scribe 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, RA and S respectively. Prove that: AB + CD = BC + DA. (2013 OD, 2016 OD)

Solution:

AP = AS ……(i) (Tangents drawn from an external point are equal in length

BP = BO …(ii)

CR = CQ ….(iii)

DR = DS ..(iv)

By adding (i) to (iv)

(AP + BP) + (CR + DR) = AS + BQ + CQ + DS

AB + CD = (BQ + CQ) + (AS + DS)

âˆ´ AB + CD = BC + AD (Hence proved)

Question 25.

In the given figure, an isosceles âˆ†ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. (2012D)

Solution:

Given: The incircle of âˆ†ABC touches the sides BC, CA and AB at D, E and F F respectively.

AB = AC

To prove: BD = CD

Proof: Since the lengths of tangents drawn from an external point to a circle are equal

âˆ´ AF = AE … (i)

BF = BD …(ii)

CD = CE …(iii)

Adding (i), (ii) and (iii), we get

AF + BF + CD = AE + BD + CE

â‡’ AB + CD = AC + BD

But AB = AC … [Given

âˆ´ CD = BD

Question 26.

In Figure, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths 8 cm and 6 cm respectively, find the value of r. (2012OD)

Solution:

Const.: Join AO, OB, CO

Proof: area of âˆ†ABC

From (i) and (ii), we get 12r = 24

âˆ´ r = 2 cm

Question 27.

In the given figure, a circle inscribed in âˆ†ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. (2013D)

Solution:

Let AD = AF = x

BD = BE = y …[Two tangents drawn from and an external point are equal

CE = CF = z

AB = 12 cm …[Given

âˆ´ x + y = 12 cm …(i)

Similarly,

y + z = 8 cm …(ii)

and x + z = 10 cm …(iii)

By adding (i), (ii) & (iii)

2(x + y + z) = 30

x + y + z = 15 …[âˆµ x + y = 12

z = 15 – 12 = 3

Putting the value of z in (ii) & (iii),

y + 3 = 8

y = 8 – 3 = 5

x + 3 = 10

x = 10 – 3 = 7

âˆ´ AD = 7 cm, BE = 5 cm, CF = 3 cm

Question 28.

The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. (2014OD)

Solution:

Given: The incircle of âˆ†ABC touches the sides BC, CA and AB at D, E and F respectively.

AB = AC

To prove: BD = CD

Proof: AF = AE ..(i)

BF = BD …(ii)

CD = CE …(iii)

Adding (i), (ii) and (iii), we get

AF + BF + CD = AE + BD + CE

â‡’ AB + CD = AC + BD

But AB = AC …[Given

âˆ´ CD = BD

Question 29.

In the figure, a âˆ†ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. If the area of âˆ†ABC is 54 cm^{2}, then find the lengths of sides AB and AC. (2011D, 2011OD, 2015 OD)

Solution:

Given: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(âˆ†ABC) = 54 cm^{2}

Joint: OA, OF, OE, OB and OC

Let AF = AE = x

BD = BF = 6 cm

CD = CE = 9 cm

âˆ´ AB = AF + BF = x + 6 …(i)

AC = AE + CE = x + 9 …(ii)

BC = DB + CD = 6 + 9 = 15 cm …(iii)

In âˆ†ABC,

Area of âˆ†ABC = 54 cm^{2} …[Given

ar(âˆ†ABC) = ar(âˆ†BOC) + ar(âˆ†AOC) + ar(âˆ†AOB)

Question 30.

In the figure, a circle is inscribed in a âˆ†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. (2016D)

Solution:

AB = 12 cm, BC = 8 cm, CA = 10 cm

As we know,

AF = AD

CF = CE

BD = BE

Let AD = AF = x cm

then, DB = AB – AD

= (12 – x) cm

âˆ´ BE = (12 – x) cm ..[Tangents drawn from an external point are equal

Similarly,

CF = CE = AC – AF = (10 â€“ x) cm

BC = 8 cm …[Given

â‡’ BE + CE = 8 â‡’ 12 – x + 10 – x = 8

â‡’ 22 – 8 = 2x â‡’ 2x = 14

âˆ´ x = 7 âˆ´ AD = x = 7 cm

BE = 12 – x = 12 – 7 = 5 cm

CF = 10 – x = 10 – 7 = 3 cm

Question 31.

In Figure, common tangents AB and CD to the two circles with lo, centres O_{1} and O_{2} intersect at E. Prove that AB = CD. (2014OD)

Solution:

EA = EC …(i) ….[Tangents drawn from an external point are equal

EB = ED …(ii)

EA + EB = EC + ED …[Adding (i) & (ii)

âˆ´ AB = CD (Hence proved)

Question 32.

If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that âˆ QPR = 120Â°, prove that 2PQ = PO. (2014D)

Solution:

âˆ OPQ = \(\frac{1}{2}\)(âˆ QPR) ..[Tangents drawn from an external point are equal

= \(\frac{1}{2}\)(120Â°) = 60Â° …[Tangent is âŠ¥ to the radius through the point of contact

âˆ OQP = 90Â°

In rt. âˆ†OQP, cos 60Â° = \(\frac{PQ}{PO}\)

\(\frac{1}{2}=\frac{P Q}{P O}\) âˆ´ 2PQ = PO

Question 33.

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQuestion (2015D)

Solution:

In âˆ†s’ TPC and TQC ….[Tangents drawn from an external point are equal

TP = TQ

TC = TC …[Common

âˆ 1 = âˆ 2 …[TP and TQ are equally inclined to OT

âˆ´ âˆ†TPC = âˆ†TQC … [SAS

âˆ´ PC = QC …[CPCT

âˆ 3 = âˆ 4 …(i)

â‡’ âˆ 3 + 24 = 180Â° … [Linear pair

â‡’ âˆ 3 + âˆ 3 = 180Â°…[From (i)

â‡’ 2âˆ 3 = 180Â° â‡’ âˆ 3 = 90Â°

âˆ´ âˆ 3 = âˆ 4 = 90Â°

âˆ´ OT is the right bisector of PQuestion

Question 34.

In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If âˆ PRQ = 120Â°, then prove that OR = PR + R. (2015OD)

Solution:

Join OP and OO

âˆ OPR = 90Â°

PR = RQ … [Tangents drawn from an external point are equal

âˆ PRO = \(\frac{1}{2}\) âˆ PRQ = \(\frac{1}{2}\) Ã— 120Â° = 60Â°

Now, In âˆ†OPR,

â‡’ âˆ OPR + âˆ POR + âˆ ORP = 180Â° …[âˆ† Rule

â‡’ 90Â° + âˆ POR + 60Â° = 180Â°

â‡’ âˆ POR + 150Â° = 180Â°

â‡’ âˆ POR = 30Â°

â‡’ sin 30Â° = \(\frac{PR}{OR}\) â‡’ \(\frac{1}{2}=\frac{P R}{O R}\)

â‡’ OR = 2PR

â‡’ OR = PR + QR (âˆµ PR = RQ) …(Hence proved)

Question 35.

In the figure, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and âˆ APB = 60Â°. Find the length of chord AB. (2016D)

Solution:

PA = PB …[Tangents drawn from an external point are equal

Given:

âˆ APB = 60Â°

âˆ PAB = âˆ PBA … (i) …(Angles opposite to equal sides

In âˆ†PAB, âˆ PAB + âˆ PBA + âˆ APB = 180Â° …[Angle-sum-property of a âˆ†

â‡’ âˆ PAB + âˆ PAB + 60Â° = 180Â°

â‡’ 2âˆ PAB = 180Â° – 60o = 120Â°

â‡’ âˆ PAB = 60Â°

â‡’ âˆ PAB = âˆ PBA = âˆ APB = 60Â°

âˆ´ APAB is an equilateral triangle

Hence, AB = AP = 5 cm …[âˆµ All sides of an equilateral A are equal

Question 36.

In the 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 âˆ OTS = âˆ OST = 30Â°. (2016OD)

Solution:

Let âˆ TOP = Î¸ …[Tangent is âŠ¥ to the radius through the point of contact

âˆ OTP = 90Â°

OT = OS = r … [Given

In rt. âˆ†OTP, cos Î¸ = \(\frac{\mathrm{OT}}{\mathrm{OP}}\)

â‡’ cos Î¸ = \(\frac{r}{2 r}\) â‡’ cos Î¸ = \(\frac{1}{2}\)

â‡’ cos Î¸ = cos 60Â° â‡’ Î¸ = 60Â°

âˆ´ âˆ TOS = 60Â° + 60Â° = 120Â°

In ATOS,

âˆ OTS = âˆ OST …[Angles opposite to equal sides

In âˆ TOS,

âˆ TOS + âˆ OTS + âˆ OST = 180Â° … [Angle-sum-property of a âˆ†

120Â° + âˆ OTS + âˆ OTS = 180Â° … [From (i)

2âˆ OTS = 180Â° – 120Â°

âˆ OTS = 60Â°/2 = 30Â°

âˆ´ âˆ OTS = âˆ OST = 30Â°

Question 37.

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 Question If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. (2017D)

Solution:

PA = PB = 12 cm …(i)

QC = AC = 3 cm …(ii)

QD = BD = 3 cm …(iii)

To find: PC + PD

= (PA – AC) + (PB – BD)

= (12 – 3) + (12 – 3) … [From (i), (ii) & (iii)

= 9 + 9 = 18 cm

Question 38.

In the figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q. (2013 OD)

Solution:

To prove: PR = RQ

Proof: PR = RC … (i)

QR = RC

From (i) and (ii), PR = QR (Hence proved)

### Circles Class 10 Important Questions Short Answer-II (3 Marks)

Question 39.

In the figure, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the lengths of QM, RN and PI. (2012OD)

Solution:

Let PL = PN = x cm

QL = QM = y cm

RN = MR = z cm

PQ = 10 cm = x + y = 10 …(i)

QR = 8 cm = y + z = 8 …(ii)

PR = 12 cm = x + z = 12 …(iii)

By adding (i), (ii) and (iii),

We get,

â‡’ 2x + 2y + 2z = 10 + 8 + 12

â‡’ 2(x + y + z) = 30

â‡’ x + y + z = 15

â‡’ 10 + z = 15 … [From (i)

âˆ´ z = 15 – 10 = 5 cm

From (ii)

y + 5 = 8

y = 8 – 5

y = 3 cm

From (iii)

x + 5 = 12

x = 12 – 5

x = 7 cm

âˆ´ QM = 3 cm, RN = 5 cm, PL = 7 cm

Question 40.

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (2012D)

Solution:

1st method:

To prove. (i) âˆ AOD + âˆ BOC = 180Â°

(ii) âˆ AOB + âˆ COD = 180Â°

Proof. In âˆ†BPO and âˆ†BQO …[Tangents drawn from an external point are equal

PO = 20 … [radii

BO = BO … [Common

âˆ†BPO = âˆ†BQO … [SSS Congruency rule

âˆ 8 = âˆ 1 …(i) (c.p.c.t.)

Similarly,

âˆ 2 = âˆ 3, âˆ 4 = âˆ 5 and âˆ 6 = âˆ 7

âˆ 1 + âˆ 2 + 23+ 24 + 25 + 26+ 27 + âˆ 8 = 360Â° …(Complete angles

âˆ 1 + âˆ 2 + 22+ 25 + 25 + 26+ âˆ 6+ âˆ 1 = 360Â°

2(âˆ 1 + âˆ 2 + 25 + 26) = 360Â°

âˆ BOC + âˆ AOD = 180Â°…(i) [Proved part I

âˆ AOB + âˆ BOC + âˆ COD + âˆ DOA = 360Â° …(Complete angles

âˆ AOB + âˆ COD + 180o = 360Â° … [From (i)

âˆ´ âˆ AOB + âˆ COD = 360Â° – 180o = 180Â° …(proved)

2nd method:

To prove:

(i) âˆ 6 + âˆ 8 = 180Â°

(ii) âˆ 5 + âˆ 7 = 180Â°

Proof. As AS and AP are tangents to the circle from a point A

âˆ´ O lies on the bisector of âˆ SAP

âˆ´ âˆ 1 = \(\frac{1}{2}\) âˆ BAD …(i)

Similarly BO, CO and DO are the bisectors of

âˆ ABC, âˆ BCD and âˆ ADC respectively. …(ii)

âˆ´ âˆ 1 + âˆ 4 + âˆ 3 + âˆ 2 =180Â°…(iii) ..[From (1) & (ii)

In âˆ†AOD, âˆ 1 + âˆ 2 + 26 = 180Â° …[Angle-sum-Prop. of a âˆ†

In âˆ†BOC, âˆ 3 + âˆ 4 + âˆ 8 = 180Â° …(v)

Adding (iv) and (v)

(âˆ 1 + âˆ 2 + 23 + 24) + 26 + 28 = 180Â° + 180Â°

180Â° + 26 + 28 = 180Â° + 180Â° … [From (iii)

âˆ´âˆ 6 + 28 = 180Â°

Now âˆ 5 + âˆ 6 + âˆ 7 + âˆ 8 = 360Â° … (Complete angles

(âˆ 5 + âˆ 7) + (âˆ 6 + âˆ 8) = 360Â°

(âˆ 5 + âˆ 7) + 180Â° = 360Â°

âˆ 5 + âˆ 7 = 360Â° – 180Â° = 180Â°

âˆ 5 + âˆ 7 = 180Â°

Question 41.

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that âˆ PTQ = 2âˆ OPQ. (2017D)

Solution:

We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Figure).

We need to prove that:

âˆ PTQ = 2âˆ OPQ

Let âˆ PTQ = Î¸

Now, TP = TQuestion ….[âˆµ Lengths of tangents drawn from an external pt. to a circle are equal

So, TPQ is an isosceles triangle.

### Circles Class 10 Important Questions Long Answer (4 Marks)

Question 42.

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (2011OD, 2012OD, 2013D, 2014OD, 2015D)

Solution:

Given: XY is a tangent at point P to the circle with centre O.

To prove: OP âŠ¥ XY

Const.: Take a point Q on XY other than P and join to OQuestion

Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.

âˆ´ OQ > OP

This happens with every point on the line XY except the point P.

OP is the shortest of all the distances of the point O to the points of XY

âˆ´ OP âŠ¥ XY … [Shortest side is âŠ¥

Question 43.

Prove that the lengths of tangents drawn from an external point to a circle are equal. (2011D, 2012OD, 2013OD, 2014, 2015D & OD

2016D & OD, 2017D)

Solution:

Given: PT and PS are tangents from an external point P to the circle with centre O.

To prove: PT = PS

Const.: Join O to P,

T & S

Proof: In âˆ†OTP and

âˆ†OSP,

OT = OS …[radii of same circle

OP = OP …[circle

âˆ OTP – âˆ OSP …[Each 90Â°

âˆ´ AOTP = AOSP …[R.H.S

PT = PS …[c.p.c.t

Question 44.

In the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP. (2014OD)

Solution:

TP = TQuestion .. [Tangents drawn from an external point

âˆ†TPQ is an isosceles âˆ† and TO is the bisector of âˆ PTQ ,

OT âŠ¥ PQ …[Tangent is âŠ¥ to the radius through the point of contact

âˆ´ OT bisects PQ

âˆ´ PR = RQ = 16 = 8 cm …[Given

In rt. âˆ†PRO,

PR^{2} + RO^{2} = PO^{2} … [Pythagoras’ theorem

8^{2} + RO^{2} = (10)^{2}

RO^{2} = 100 – 64 = 36

âˆ´ RO = 6 cm

Let TP = x cm and TR = y cm

Then OT = (y + 6) cm

In rt. âˆ†PRT, x^{2} = y^{2} + 8^{2} …(i) …[Pythagoras’ theorem

In rt. âˆ†OPT,

OT^{2} = TP^{2} + PO^{2} …(Pythagoras’ theorem

(y + 6)^{2} = x^{2} + 10^{2}

y^{2} + 12y + 36 = y^{2} +64 + 100 …[From (i)

12y = 164 – 36 = 128 â‡’ y = \(\frac{128}{12}=\frac{32}{3}\)

Putting the value of y in (i),

Question 45.

In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that âˆ RPQ = 30Â°. A chord RS is drawn parallel to the tangent PQuestion Find âˆ RQS. (2015D)

Solution:

PR = PO …[âˆµ Tangents drawn from an external point are equal

â‡’ âˆ PRQ = âˆ PQR …[âˆµ Angles opposite equal sides are equal

In âˆ†PQR,

â‡’ âˆ PRQ + âˆ RPQ + âˆ POR = 180Â°…[âˆ† Rule

â‡’ 30Â° + 2âˆ PQR = 180Â°

â‡’ \(\angle \mathrm{PQR}=\frac{(180-30)^{\circ}}{2}\)

= 75Â°

â‡’ SR || QP and QR is a transversal

âˆµ âˆ SRQ = âˆ PQR … [Alternate interior angle

âˆ´ âˆ SRO = 75Â° …..[Tangent is I to the radius through the point of contact

â‡’ âˆ ORP = 90Â°

âˆ´ âˆ ORP = âˆ ORQ + âˆ QRP

90Â° = âˆ ORQ + 75Â°

âˆ ORQ = 90Â° – 75o = 150

Similarly, âˆ RQO = 15Â°

In âˆ†QOR,

âˆ QOR + âˆ QRO + âˆ OQR = 180Â° …[âˆ† Rule

âˆ´âˆ QOR + 15Â° + 15Â° = 180Â°

âˆ QOR = 180Â° – 30Â° = 150Â°

â‡’ âˆ QSR = \(\frac{1}{2}\)âˆ QOR

â‡’ âˆ QSR = \(\frac{150^{\circ}}{2}\) = 750 … [Used âˆ SRQ = 75Â° as solved above

In ARSQ, âˆ RSQ + âˆ QRS + âˆ RQS = 180Â° … [âˆ† Rule

âˆ´ 75Â° + 75Â° + âˆ RQS = 180Â°

âˆ RQS = 180Â° – 150o = 30Â°

Question 46.

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. (2015OD)

Solution:

B is the mid point of arc (ABC)

OA = OC …[Radius

OF = OF …[Common

âˆ´ âˆ 1 = âˆ 2 …[Equal angles opposite equal sides

âˆ´ âˆ†OAF = âˆ†OCF (SAS)

âˆ´ âˆ AFO = âˆ CFO = 90Â° …[c.p.c.t

â‡’ âˆ AFO = âˆ DBO = 90Â° …[Tangent is âŠ¥to the radius through the point of contact

But these are corresponding angles,

âˆ´ AC || DE

Question 47.

In the 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. (2016D)

Solution:

âˆ OPT = 90Â° …[Tangent is âŠ¥ to the radius through the point of contact

We have, OP = 5 cm, OT = 13 cm

In rt. âˆ†OPT,

OP^{2} + PT^{2} = OT? …[Pythagoras’ theorem

â‡’ (5)^{2} + PT^{2} = (13)^{2}

â‡’ PT^{2} = 169 – 25 = 144 cm

â‡’ PT = \(\sqrt{144}\)

= 12 cm

OP = OQ = OE = 5 cm … [Radius of the circle

ET = OT – OE

= 13 – 5 = 8 cm

Let, PA = x cm, then AT = (12 – x) cm

PA = AE = x cm …[Tangent drawn from an external point

In rt. âˆ†AET,

AE^{2} + ET^{2} = AT^{2} …(Pythagoras’ theorem

â‡’ x^{2} + (8)^{2} = (12 – x)^{2}

â‡’ x^{2} + 64 = 144 + x^{2} – 24x

â‡’ 24x = 144 – 64

x = \(\frac{80}{24}=\frac{10}{3}\) cm

AB = AE + EB = AE + AE = 2AE = 2x :

âˆ´ AB = \(2\left(\frac{10}{3}\right)=\frac{20}{3} \mathrm{cm}=6 \frac{2}{3}\) cm

or 6.67 cm or 6.6 cm

Question 48.

In the figure, two equal circles, with centres 0 and O’, touch each other A 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 \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\). (2016OD)

Solution:

Given: Two equal circles, with centres O and Oâ€™, touch each other at point X. OO’ is produced to meet the circle with centre (at A. AC is tangent to the circle with centre O, at the point C. OÊ»D is perpendicular to AC.

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

Proof: âˆ ACO = 90Â° … [Tangent is âŠ¥ to the radius through the point of contact

In âˆ†AO’D and âˆ†AOC

âˆ Oâ€™AD = âˆ OAC …(Common

âˆ´ âˆ ADO = âˆ ACO …[Each 90Â°

âˆ´ âˆ†AO’D ~ âˆ´AOC …(AA similarity

\(\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}=\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) … [In ~ As corresponding sides are proportional

\(\frac{r}{3 r}=\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) …[Let AO’ = O’X = OX = r â‡’ AO = r +r+ r = 3r

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

Question 49.

In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that âˆ DOE = 90Â°. (2013D)

Solution:

Proof: Let I be XY and m be XY’

âˆ XDE + âˆ X’ED = 180Â° … [Consecutive interior angles

\(\frac{1}{2}\)XDE + \(\frac{1}{2}\)âˆ Xâ€™ED =

= \(\frac{1}{2}\) (180Â°)

= âˆ 1 + âˆ 2 = 90Â° …[OD is equally inclined to the tangents

In âˆ†DOE, âˆ 1 + âˆ 2 + 23 = 180Â° …[Angle-sum-property of a âˆ†

90Â° + 23 = 180Â°

â‡’ âˆ 3 = 180Â° – 90o = 90Â°

âˆ´ âˆ DOE = 90Â° …(proved)

Question 50.

In the figure, the sides AB, BC and CA of triangle ABC touch a circle with centre o and radius r at P, Q and R. respectively. (2013OD)

Prove that:

(i) AB + CQ = AC + BQ

(ii) Area (AABC) = \(\frac{1}{2}\) (Perimeter of âˆ†ABC ) Ã— r

Solution:

Part I:

Proof: AP = AR …(i)

BP = BQ … (ii)

CQ = CR … (iii)

Adding (i), (ii) & (iii)

AP + BP + CQ

= AR + BQ + CR

AB + CQ = AC + BQ

Part II: Join OP, OR, OQ, OA, OB and OC

Proof: OQ âŠ¥ BC; OR âŠ¥ AC; OP âŠ¥ AB

ar(âˆ†ABC) = ar(âˆ†AOB) + ar(âˆ†BOC) + ar (âˆ†AOC)

Area of (âˆ†ABC)

Question 51.

In the figure, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC. (2014OD)

Solution:

Let AE = x âˆ´ AF = x

BC = 8 + 6 = 14 cm

AB = (x + 8) cm

AC = (x + 6) cm

âˆ 1 = âˆ 2 = 23 = 90Â° …[ Tangent is âŠ¥ to the radius B [through the point of contact

\(4 \sqrt{3 x(x+14)}\) = 2(2x + 28)

\(4 \sqrt{3 x(x+14)}\) = 2.2(x + 14)

3x(x + 14) = (x + 14)^{2} … [Squaring both sides

3x(x + 14) – (x + 14)^{2} = 0

(x + 14) [3x – (x + 14)] = 0

(x + 14) (2x – 14) = 0

x = -14 or x = 7

âˆ´ x = 7 … [As side of âˆ† cannot be -ve

âˆ´ AB = x + 8 = 15 cm

and AC = x + 6 = 13 cm

Question 52.

Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre. (2014 D)

Solution:

Given: CD and EF are two C parallel tangents at points A and B of a circle with centre O.

To prove: AB passes through centre O or AOB is diameter of the circle.

Const.: Join OA and OB. Draw OM || CD.

Proof: âˆ 1 = 90Â° … (i)

…[âˆµ Tangent is I to the radius through the point of contact

OM || CD

âˆ´ âˆ 1 + âˆ 2 = 180Â° …(Co-interior angles

90Â° + âˆ 2 = 180Â° …[From (i)

âˆ 2 = 180Â° – 90o = 90Â°

Similarly, âˆ 3 = 90Â°

âˆ 2 + âˆ 3 = 90Â° + 90Â° = 180Â°

âˆ´ AOB is a straight line.

Hence AOB is a diameter of the circle with centre O.

âˆ´ AB passes through centre 0.