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SEBA Class 10 Mathematics Chapter 10 Circles
Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 10 Circles Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 10 Circles Solutions for All Subject, You can practice these here.
Circles
Chapter – 10
| Exercise 10.1 |
1. How many tangents can a circle have?
Ans: A circle can have infinitely many tangents.
2. Fill in the blanks:
(i) A tangent to a circle intersects it in……
Ans: A tangent to a circle intersects it in one point (s).
(ii) A line intersecting a circle in two points is called a parallel tangent at the most…….. point (s).
Ans: A line intersecting a circle in two points is called a secant.
(iii) A circle can have ….
Ans: A circle can have two parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called
Ans: The common point of a tangent to a circle and the circle is called point of contact.
3. Tangent PQ at a point P of a circle of radius 5cm meets a line through the centre O at a point Q so that OQ = 12cm Length PQ is:
(a) 12cm
(b) 13cm
(c) 8.5 cm
(d) sqrt(119)
Ans: (d) sqrt(119)
4. Draw a circle and and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.
| Exercise 10.2 |
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
Ans:
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 ∆OPQ, 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
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
Ans: It is given that TP and TQ are tangents.
Therefore, the 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
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
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 and ∠OAP = 90º
In AOBP,
Sum of all interior angles = 360
∠OAP + ∠APB +∠PBO + ∠BOA = 360
90º + 80º +90º + ∠BOA = 360
∠BOA = 100
In ∆OPB and ∆OPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ∆OPB ≅ ∆OPA (SSS congruence criterion)
And thus, ∠POB = ∠POA
Hence, alternative (A) is correct.
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB be the 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
くOAR = 90 O く OAS
= 90 O
くOBP = 90 O
く OBQ = 90 O
It can be observed that
くOAR = く OBQ ( Alternate interior angles)
くOAS = く ( Alternate interior angles)
Since alternate interior angles)
Since alternative interior angles are equal , lines PQ and RS Will parallel.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Ans: 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 passes through o, therefore
く OPB = 90 ………………………………………..(1)
O Is the centre of tye circle and p is the points of contrat. We know the line joining the centre and the points of contract to the tangent of the circle perpendicular to each other.
∴ くOPB = 90 ………………………(2)
Comparing equations (1) and (2) we obtain
く O’PB = くOPB………………………………(3)
From the figure , it can be observed that
くOBP = く OPB…………………………….(4)
Therefore , く OPB = く OPB is not possible. It is only possible , when the line O’P Coincides with op
Therefore , the perpendicular to ab through centre O
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.
Ans:
let us consider a circle centred at point O.
AB is a tangent drawn on this circle from point A.
Given that,
OA = 5cm and AB = 4cm
In Δ ABO,
OB ⊥ AB ( radius ⊥ tangent at at the point of Contact)
Applying pythagoras theorem in Δ ABO,
OB 丄 AB ( radius 丄 tangent at the points of contract )
Applying pythagoras theorem in Δ ABO, we obtain
AB2 + BO2 = OA2
42 + BO2 = 25
16 + BO2 = 25
BO2 = 3
Hence , the radius of the circle is 3 cm.
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.
Ans:
Let the two concentric circles be a contread at point o . and let PQ be the chord of the larger circle which touches the smaller circle at point. Therefore PQ is tangent to the smaller circle at the point.
OA 丄 PQ (As oa is the circle)
Applying pythagoras theorem in Δ oap , we obtain)
OA ⊥ PQ ( As + AP2 = OP2
32 + AP2 + = 52
AP2 = 16
AP = 4
In ΔOPQ,
Since OA ⊥ PQ
Ap = AQ (Perpendicular from the centre of the circle bisects chord)
∴ PQ = 2AP = 2 χ 45 = 8
Therefore , the chord of the larger circle is 8 cm .
8. A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that AB + CD = AD + BC
Ans: 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
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.
Ans: let join point o to c.
In Δ OPA and Δ OCA ,
OP = Δ OPA and Δ OCA,
OP = OC ( Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
ΔOPA ≅ Δ OCA (SSS congruence criterion )
くPOA = く COA ………………………. (i)
Similarly , Δ OQB ≅ ΔOCB
く QOB = くCOB …………………………..(ii)
Since POQ is a diameter of the circle , itis straight line.
Therefore , くPOQ + くCOA + くCOB + く QOB = 180O
From equations (i) and (ii) , it can be observed that
2 く COQ + 2 く COB = 180O
2く COA + 2 く COB = 90O
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.
Ans:
Let us consider a circle centred 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 centre 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.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Ans:
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.
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.
Ans: 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
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
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Ans:
Let ABCD be a quadrilateral circumscribing a circle centred 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 centre of the circle.Consider ∆OAP and ∆OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the same circle)
OA = OA (Common side)
∆OAP ≅ ∆OAS (SSS congruence criterion)
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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