Page No 204:
Question 1:
How many tangents can a circle have?
Answer:
A circle can have infinite tangents.
Question 2:
Fill in the blanks:
(i) A tangent to a circle intersects it in _______ point (s).
(ii) A line intersecting a circle in two points is called a __________.
(iii) A circle can have __________ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ____.
Answer:
(i) One
(ii) Secant
(iii) Two
(iv) Point of contact
Question 3:
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm. (B) 13 cm (C) 8.5 cm (D) 
cm test
Answer:
We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
OP 
PQ
By applying Pythagoras theorem in ΔOPQ,
OP2 + PQ2 = OQ2
52 + PQ2 =122
PQ2 =144 − 25
PQ = cm.
Hence, the correct answer is (D).
Question 4:
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
It can be observed that AB and CD are two parallel lines. Line AB is intersecting the circle at exactly two points, P and Q. Therefore, line AB is the secant of this circle. Since line CD is intersecting the circle at exactly one point, R, line CD is the tangent to the circle.
Page No 209:
Question 3:
Answer:
Video Solution
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.
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.
Video Solution
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.
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.
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