RBSE Solutions for Class 10 Maths Chapter 13 Circle and Tangent Ex 13.1

RBSE Solutions for Class 10 Maths Chapter 13 Circle and Tangent Ex 13.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 13 Circle and Tangent Exercise 13.1.

Rajasthan Board RBSE Class 10 Maths Chapter 13 Circle and Tangent Ex 13.1

Ex 13.1 Class 10 RBSE Question 1.
Write true or false. Also (RBSESolutions.com) write the reason for your answer.
(i) Tangent of circle is the line which intersect the circle at two points.
(ii) A tangent line XY, touches a circle at point P whose center is O and Q tangent line is at different point, then OP = OQ.
(iii) The points situated on the circle are P and Q has two tangent lines LM and XY. If PQ is diameter, then LM || XY.
(iv) The center of circle ‘O’ is situated on the other circle whose centre is A. If circle having center passes from A and B in such a way that AOB are on a line then the tangent line drawn from B will passes through the intersecting points of both circles.
Solution :
(i) False : There is one and only one tangent passing through a point lying on a circle.
(ii) False : Since OP is perpendicular to tangent and perpendicular is smaller than other distance.
(iii) True: Since tangent is perpendicular to diameter.
(iv) True : Since AOB is a diameter and angle made in semi circle is right angle.

RBSE Solutions For Class 10 Maths Chapter 13.1 Question 2.
Fill in the blanks:
(i) ……… tangent lines can be (RBSESolutions.com) drawn from a point situated on circle.
(ii) A line which intersects the circle at two points is known as ……..
(iii) A circle can have ………. parallel tangent lines.
(iv) The common point of tangent line and circle ¡s known as ……….
Solution :
(i) One
(ii) Secant
(iii) Two
(iv) point of contact

RBSE Class 10 Maths Chapter 13 Question 3.
Two concentric circles have radius 5 cm and 3 cm respectively. Find the length of chord of circle which touches the smaller circle.
Solution :
Let two concentric circle (RBSESolutions.com) with center O.
Let radius OA and OP are 5 cm and 3 cm respectively.
AB is the chord of bigger circle which touches smaller circle at point P.
∴ OP ⊥ AB (By theorem 13.1)
∠OPA = 90°

∴ In right angled ∆OPA, by Pythagoras theorem.
AP2 + OP2 = OA2
⇒ AP2 + (3)2 = (5)2
⇒ AP2 = (5)2 – (3)2
= 25 – 9 = 16
∴ AP = 4 cm
But in bigger circle, OP is (RBSESolutions.com) perpendicular to chord AB from center O.
∴ Point P bisects the chord AB.
AP = BP = 4 cm
Length of chord AB = AP + BP
= 4 + 4 = 8 cm
So, Length of chord of bigger circle = 8 cm.

RBSE Solutions For Class 10 Maths Chapter 13 Question 4.
If the length of tangent line is 4 cm drawn from any point, 10 cm away from the center of circle, then what will be the radius of that circle ?
Solution :
PQ is the tangent drawn an circle (RBSESolutions.com) from external point Q and OP is radius of circle.

So ∠OPQ = 90° (by theorem 13.1)
From ∆OPQ
OQ2 =PQ2 + OP2 (By Pythagoras theorem)
102 = 42 + OP2
OP2 = 102 – 42
= 100 – 16
= 84
OP = $$\sqrt { 84 }$$ cm
OP = 2$$\sqrt { 21 }$$ cm

Ex 13.1 Class 10 Question 5.
A circle with center ‘O’ touches all the four sides of a quadrilateral ABCD internally in such a way that AB is divided in the ratio 3 : 1 and AB = 8 cm, then find the radius of circle if OA = 10 m.
Solution :
A circle with center O touches (RBSESolutions.com) the four side of a quadrilateral ABCD of the point E, F, G and H respectively. Contact point E divides side AB in the ratio 3 : 1.

OA = 10 cm
AB = 8 cm
Let AE = 3x
EB = x
∴ AB = AE + EB
⇒ 8 = 3x + x
⇒ 8 = 4x
⇒ x = 2
AE = 3 × 2 = 6 cm
∴ EB = x = 2 cm
From right angled ∆AEO
OA2 = AE2 + OE2
(10)2 = (6)2 + (OE)2
(OE)2 = (10)2 – (6)2
= 100 – 36 = 64
OE = $$\sqrt { 64 }$$
OE = 8 cm
Hence, radius of circle = 8 cm

Class 10th Maths Chapter 13 Exercise 13.1 Question 6.
A circle touches all the sides (RBSESolutions.com) of a quadrilateral. Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Solution :
Given :
A quadrilateral PQRS circumscribes a circle with center O. Sides of quadrilateral PQ, QR, RS and SP touches the circle at point L, M, N, T respectively.

To prove : ∠POQ + ∠SOR = 180°
and ∠SOP + ∠ROQ = 180°.
Construction : Join P, Q, R, L, M, N and T with center O of circle.
Proof: Since, OL, OM, ON and OT are radius (RBSESolutions.com) of circle and QL, MQ, RN and ST are tangents of circle. So
QL ⊥ OL, QM ⊥ OM, RN ⊥ ON and ST ⊥ OT
Now in right angled ΔOMQ and right ΔOLQ
∠OMQ = ∠OLQ (each 90°)
hypotenuse OQ = hypotenuse OQ (common side)
and OM = OL (equal radii of circle)
∴ OMQ = OLQ (By RHS Congruence)
⇒ ∠3 = ∠2 (CPCT)
Similarly ∠4 = ∠5
∠6 = ∠7 and ∠8 = ∠1
∵ Sum of all angles made on point O of center of circle = 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ ∠1 + ∠2 + ∠2 + ∠5 + ∠5 + ∠6 + ∠6 + ∠1 = 360°
⇒ 2(∠1 + ∠2 + ∠5 + ∠6) = 360°
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180°
∠POQ + ∠SOR = 180°
[ ∵∠1 + ∠2 = ∠POQ and ∠5 + ∠6 = ∠SOR ]
Similarly ∠SOP + ∠ROQ = 180°
So, opposite sides of a quadrilateral (RBSESolutions.com) circumscribing a circle subtend supplementary angles at the center of circle.

Class 10 Math Ch 13 Ex 13.1 Question 7.
In following figure, center of a circle is O and tangent drawn from point P, PA and PB which touches the circle at A and B respectively, then prove that OP line segment is bisector of AB.

Solution :
Let OP intersect chord AB at point C.
To prove : AC = BC.
Proof : In ΔPCA and ΔPCB
PA = PB (tangents at circle (RBSESolutions.com) from external point P.)
PC = PC (common side)
∠APC = ∠BPC [∵ tangents PA and PB, make equal angle with OP]
by SAS congruence,
ΔPCA = ΔPCB
⇒ AC = BC
Hence, OP bisects line segment AB.

Class 10 Chapter 13.1 Question 8.
In following figure from exterior point ‘P’ two tangent line PA and PB touches the circle at A and B, then prove that PAOB is a cyclic quadrilateral where O is the (RBSESolutions.com) center of circle.

Solution :
We know that tangent at any point of a (RBSESolutions.com) circle is perpendicular to radius through the point of contact.
OA ⊥ PA and OB ⊥ PB
∴ ∠OAP = ∠OBP = 90°
⇒ ∠OAP + ∠OBP = 90° + 90° = 180°