Circles NCERT Solution, study material, CBSE Notes

NCERT Solution: Circles

How many tangents can a circle have?

A circle can have infinite tangents.

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 ............

(i) one

(ii) secant

(iii) two

(iv) point of contact

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) √119 cm

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,

OQ² = OP² +PQ²
⇒ (12)²= 5² + PQ²

⇒PQ² = 144 - 25

⇒PQ² = 119

⇒PQ = √119 cm

(D) is the correct option.

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

AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.

choose the correct option and give justification.

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

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem in ΔOPQ,

OQ² = OP² +PQ²
⇒ (25)²= OP² + (24)²

⇒ OP² = 625 - 576

⇒ OP² = 49

⇒ OP = 7 cm

The radius of the circle is option (A) 7 cm.

choose the correct option and give justification.

In Fig. 10.11, 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°

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

∠PTQ is equal to option (B) 70°.

choose the correct option and give justification.

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

OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°

In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°

⇒ ∠AOB = 100°

Now,

In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)

Thus ∠POB = ∠POA

∠AOB = ∠POB + ∠POA

⇒ 2 ∠POA = ∠AOB

⇒ ∠POA = 100°/2 = 50°

∠POA is equal to option (A) 50°

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

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,

∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.

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

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