Circles | Exercise 10.2

Question 2

In Q.1 to 3, choose the correct option and give justification.

  1. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that POQ=110\angle \text{POQ} = 110^\circ, then PTQ\angle \text{PTQ} is equal to (A) 6060^\circ (B) 7070^\circ (C) 8080^\circ (D) 9090^\circ
Question diagram 1
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Solution
Understand the Question
  • The tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, OPT=90\angle \text{OPT} = 90^\circ and OQT=90\angle \text{OQT} = 90^\circ.
  • In quadrilateral OPTQ\text{OPTQ}, the sum of interior angles is 360360^\circ.
  • Since the two angles at the points of contact sum to 180180^\circ, the central angle POQ\angle \text{POQ} and the angle between the tangents PTQ\angle \text{PTQ} are supplementary: POQ+PTQ=180\angle \text{POQ} + \angle \text{PTQ} = 180^\circ.

Step 1 · Find Angles at the Points of Contact

Question diagram

Since the radius is perpendicular to the tangent at the point of contact: OPTP    OPT=90\text{OP} \perp \text{TP} \implies \angle \text{OPT} = 90^\circ OQTQ    OQT=90\text{OQ} \perp \text{TQ} \implies \angle \text{OQT} = 90^\circ

Step 2 · Calculate PTQ\angle \text{PTQ} using Angle Sum Property

In quadrilateral OPTQ\text{OPTQ}, the sum of all interior angles is 360360^\circ: POQ+OQT+PTQ+OPT=360\angle \text{POQ} + \angle \text{OQT} + \angle \text{PTQ} + \angle \text{OPT} = 360^\circ

Substitute the known values: 110+90+PTQ+90=360110^\circ + 90^\circ + \angle \text{PTQ} + 90^\circ = 360^\circ

290+PTQ=360290^\circ + \angle \text{PTQ} = 360^\circ

PTQ=360290=70\angle \text{PTQ} = 360^\circ - 290^\circ = 70^\circ

Answer

(B) 7070^\circ

Common Mistakes
  • Direct Supplementary Shortcut: Forgetting that the angle subtended by the tangents (PTQ\angle \text{PTQ}) and the central angle (POQ\angle \text{POQ}) are directly supplementary: PTQ=180110=70\angle \text{PTQ} = 180^\circ - 110^\circ = 70^\circ.
  • Tangent-Radius Angle Error: Assuming the angle between the tangent and radius is something other than 9090^\circ.

More questions in Exercise 10.2

Q1

In Q.1 to 3, choose the correct option and give justification.

  1. From a point QQ, the length of the tangent to a circle is 24 cm24\text{ cm} and the distance of QQ from the centre is 25 cm25\text{ cm}. The radius of the circle is (A) 7 cm7\text{ cm} (B) 12 cm12\text{ cm} (C) 15 cm15\text{ cm} (D) 24.5 cm24.5\text{ cm}
Q2

In Q.1 to 3, choose the correct option and give justification.

  1. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that POQ=110\angle \text{POQ} = 110^\circ, then PTQ\angle \text{PTQ} is equal to (A) 6060^\circ (B) 7070^\circ (C) 8080^\circ (D) 9090^\circ
Q3

In Q.1 to 3, choose the correct option and give justification.

  1. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 8080^\circ, then POA\angle \text{POA} is equal to (A) 5050^\circ (B) 6060^\circ (C) 7070^\circ (D) 8080^\circ
Q4

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

Q5

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Q6

The length of a tangent from a point AA at distance 5 cm5\text{ cm} from the centre of the circle is 4 cm4\text{ cm}. Find the radius of the circle.

Q7

Two concentric circles are of radii 5 cm5 \text{ cm} and 3 cm3 \text{ cm}. Find the length of the chord of the larger circle which touches the smaller circle.

Q8

A quadrilateral ABCDABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that

AB+CD=AD+BC\text{AB} + \text{CD} = \text{AD} + \text{BC}

Q9

In Fig. 10.13, XYXY and XYX'Y' are two parallel tangents to a circle with centre OO and another tangent ABAB with point of contact CC intersecting XYXY at AA and XYX'Y' at BB. Prove that AOB=90\angle AOB = 90^\circ.

Q10

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.

Q11

Prove that the parallelogram circumscribing a circle is a rhombus.

Q12

A triangle ABCABC is drawn to circumscribe a circle of radius 4 cm4 \text{ cm} such that the segments BDBD and DCDC into which BCBC is divided by the point of contact DD are of lengths 8 cm8 \text{ cm} and 6 cm6 \text{ cm} respectively (see Fig. 10.14). Find the sides ABAB and ACAC.

Q13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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