NCERT Class 10 Mathematics Chapter 10 Circles Solutions, NCERT Solutions For Class 10 Maths, CBSE Solutions For Class 10 Mathematics to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 10 Mathematics Chapter 10 Circles Notes and select needs one.
NCERT Class 10 Mathematics Chapter 10 Circles
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 10 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT 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 _______________ point (s).
Ans: A tangent to a circle intersects it in one point (s).
(ii) A line intersecting a circle in two points is called ________________.
Ans: A line intersecting a circle in two points is called a secant.
(iii) A circle can have _______________ parallel tangents at the most.
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.
Ans:
| 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: (A) Reason: angle OPT = 90 deg angle TQO = 90 deg In Quad OPTQ Tangent is perpendicular to the tangent through the point of contact. angle OPT + angle PTQ+ angle TQO + angle POQ = 360 deg [Sum of all angles o quad. is 360°]
90^ + angle PTQ + 90 deg + 110 deg = 360 deg
angle PTQ = 360 deg – 110 deg – 90 deg – 90 deg
.. angle PTQ = 70
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
Ans: 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.
Ans: 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
くOAR = 90° くOAS
= 90°
くOBP = 90°
く OBQ = 90°
It can be observed that
くOAR =くOBQ ( Alternate interior angles)
くOAS =く(Alternate interior angles)
(Since alternate interior angles)
Since alternatives 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 the 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:
∠OPA = 90° [Tangent is to the radius through the pt. of Contact]
In rt ∆OPA, OP + PA2 = OA2 [Pythagoras theorem]
OP² + 42=52
OP² = 25-16 = 9
Radius OP = √9 = 3
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.
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.
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:
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Ans:
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.
