NCERT solution class 10 chapter 10 Circles exercise 10.2 mathematics

EXERCISE 10.2

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Theorem, Ex. 10.2, Circles, Class 10th Maths

Q 1, Ex. 10.2, Page No 213, Circles, NCERT Solutions, Class 10th Mathematics

Page No 213:

Question 1:

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm
(C) 15 cm (D) 24.5 cm

Answer:

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ⊥ PQ
Applying Pythagoras theorem in ΔOPQ, we obtain
OP² + PQ² = OQ²
OP² + 24² = 25²
OP² = 625 − 576
OP² = 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct.

Q 2, Ex. 10.2, Page No. 213, Circles, NCERT Solutions, Class 10th Maths

Question 2:

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110, then ∠PTQ is equal to
(A) 60 (B) 70
(C) 80 (D) 90

Answer:

It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ⊥ TP and OQ ⊥ TQ
∠OPT = 90º
∠OQT = 90º
In quadrilateral POQT,
Sum of all interior angles = 360
∠OPT + ∠POQ +∠OQT + ∠PTQ = 360
⇒ 90+ 110º + 90 +PTQ = 360
⇒ PTQ = 70
Hence, alternative (B) is correct.

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Q 3, Ex. 10.2, Page No. 213, Circles, NCERT Solutions, Class 10th Maths

Question 3:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80, then ∠POA is equal to
(A) 50 (B) 60
(C) 70 (D) 80

Answer:

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ PA and OB ⊥ PB
∠OBP = 90º
∠OAP = 90º
In AOBP,
Sum of all interior angles = 360
∠OAP + ∠APB +∠PBO + ∠BOA = 360
90 + 80 +90º +BOA = 360
∠BOA = 100
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)
A ↔ B, P ↔ P, O ↔ O
And thus, ∠POB = ∠POA

Hence, alternative (A) is correct.

Q 4, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Page No 214:

Question 4:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.
Radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = 90º
∠OAS = 90º
∠OBP = 90º
∠OBQ = 90º
It can be observed that
∠OAR = ∠OBQ (Alternate interior angles)
∠OAS = ∠OBP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.

Q 5, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 5:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,
∠O’PB = 90° … (1)
O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.
∴ ∠OPB = 90° … (2)
Comparing equations (1) and (2), we obtain
∠O’PB = ∠OPB … (3)
From the figure, it can be observed that,
∠O’PB < ∠OPB … (4)
Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.
Therefore, the perpendicular to AB through P passes through centre O.

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Q 6, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 6:

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:

Let us consider a circle centered at point O.
AB is a tangent drawn on this circle from point A.
Given that,
OA = 5cm and AB = 4 cm
In ΔABO,
OB ⊥ AB (Radius ⊥ tangent at the point of contact)
Applying Pythagoras theorem in ΔABO, we obtain
AB² + BO² = OA²
4² + BO² = 5²
16 + BO² = 25
BO² = 9
BO = 3
Hence, the radius of the circle is 3 cm.

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Q 7, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 7:

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.
OA ⊥ PQ (As OA is the radius of the circle)
Applying Pythagoras theorem in ΔOAP, we obtain
OA² + AP² = OP²
3² + AP² = 5²
9 + AP² = 25
AP² = 16
AP = 4
In ΔOPQ,
Since OA ⊥ PQ,
AP = AQ (Perpendicular from the center of the circle bisects the chord)
PQ = 2AP = 2 × 4 = 8
Therefore, the length of the chord of the larger circle is 8 cm.

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Q 8, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 8:

A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that AB + CD = AD + BC

Answer:

It can be observed that
DR = DS (Tangents on the circle from point D) … (1)
CR = CQ (Tangents on the circle from point C) … (2)
BP = BQ (Tangents on the circle from point B) … (3)
AP = AS (Tangents on the circle from point A) … (4)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC

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Q 9, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 9:

In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB=90.

Answer:

Let us join point O to C.

In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
ΔOPA ΔOCA (SSS congruence criterion)
Therefore, P ↔ C, A ↔ A, O ↔ O
∠POA = ∠COA … (i)
Similarly, ΔOQB ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º
From equations (i) and (ii), it can be observed that
2∠COA + 2 ∠COB = 180º
∠COA + ∠COB = 90º
∠AOB = 90°

Q 10, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 10:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer:

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA (radius) ⊥ PA (tangent)
Therefore, ∠OAP = 90°
Similarly, OB (radius) ⊥ PB (tangent)
∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB+∠PBO +∠BOA = 360º
90º + ∠APB + 90º + ∠BOA = 360º
∠APB + ∠BOA = 180º
Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Q 11, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 11:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Since ABCD is a parallelogram,
AB = CD …(1)
BC = AD …(2)

It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.

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Q 12, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 12:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.
In ABC,
CF = CD = 6cm (Tangents on the circle from point C)
BE = BD = 8cm (Tangents on the circle from point B)
AE = AF = x (Tangents on the circle from point A)
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
s = 14 + x

Area of ΔOBC = 
Area of ΔOCA =
Area of ΔOAB =
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Either x+14 = 0 or x − 7 =0
Therefore, x = −14and 7
However, x = −14 is not possible as the length of the sides will be negative.
Therefore, x = 7
Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm

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Q 13, Ex. 10.2, Page No. 214, Circles, NCERT Solutions, Class 10th Maths

Question 13:

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

Answer:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.
Consider ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the same circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence criterion)
Therefore, A ↔ A, P ↔ S, O ↔ O
And thus, ∠POA = ∠AOS
∠1 = ∠8
Similarly,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º
(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º
2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
(∠1 + ∠2) + (∠5 + ∠6) = 180º
∠AOB + ∠COD = 180º
Similarly, we can prove that ∠BOC + ∠DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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