Circles | Exercise 10.2

Question 9

In Fig. 10.13, XY and X′Y′\text{X}'\text{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′\text{X}'\text{Y}' at B. Prove that ∠AOB=90∘\angle\text{AOB} = 90^\circ.

Question diagram 1
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Solution

Parallel Tangents: Two tangents to a circle that are parallel to each other. They always touch the circle at opposite ends of a diameter, so the line joining their points of contact (PQ) passes through the centre O.

Tangents from an External Point are Equal: Two tangents drawn from the same external point to a circle are always equal in length. Used here for point A (tangents AP and AC) and point B (tangents BQ and BC).

We will use properties of tangents and congruent triangles to prove the required angle.

Step 1 — Congruence of △OPA\triangle OPA and △OCA\triangle OCA

Let's join point C to the center O. Consider △OPA\triangle OPA and △OCA\triangle OCA. OP and OC are radii of the same circle. So, OP=OCOP = OC. AP and AC are tangents from point A. Tangents from an external point are equal. So, AP=ACAP = AC. AO is a common side to both triangles. So, AO=AOAO = AO. SSS Congruence Rule: If all three sides of one triangle equal the corresponding three sides of another, the triangles are congruent.

By SSS congruence rule, △OPA≅△OCA\triangle OPA \cong \triangle OCA.

CPCT (Corresponding Parts of Congruent Triangles): All corresponding angles and sides of congruent triangles are equal.

This means their corresponding angles are equal.

∠POA=∠COA(1)\boxed{\angle POA = \angle COA \quad (1)}

Diagram 1

Step 2 — Congruence of △OQB\triangle OQB and △OCB\triangle OCB

Similarly, consider △OQB\triangle OQB and △OCB\triangle OCB. OQ and OC are radii of the same circle. So, OQ=OCOQ = OC. BQ and BC are tangents from point B. Tangents from an external point are equal. So, BQ=BCBQ = BC. BO is a common side to both triangles. So, BO=BOBO = BO. By SSS congruence rule, △OQB≅△OCB\triangle OQB \cong \triangle OCB. This means their corresponding angles are equal.

∠QOB=∠COB(2)\boxed{\angle QOB = \angle COB \quad (2)}

Step 3 — Calculate ∠AOB\angle AOB

Why POQ is a straight line: XY and X'Y' are parallel tangents touching the circle at P and Q. The line joining opposite points of tangency of parallel tangents always passes through the centre — making POQ a diameter and hence a straight line.

Angles on a Straight Line: Angles that together form a straight line always sum to 180°.

XY and X'Y' are parallel tangents. The line segment PQ passes through the center O. So, POQ is a straight line. Angles on a straight line sum to 180∘180^\circ. So, ∠POQ=180∘\angle POQ = 180^\circ. We can write ∠POQ\angle POQ as a sum of four angles. ∠POQ=∠POA+∠COA+∠COB+∠QOB\angle POQ = \angle POA + \angle COA + \angle COB + \angle QOB. Substitute the results from (1) and (2). We have ∠POA=∠COA\angle POA = \angle COA and ∠QOB=∠COB\angle QOB = \angle COB.

2∠COA+2∠COB=180∘2\angle COA + 2\angle COB = 180^\circ

Divide the entire equation by 2.

∠COA+∠COB=90∘\angle COA + \angle COB = 90^\circ

The sum of ∠COA\angle COA and ∠COB\angle COB forms ∠AOB\angle AOB.

∠AOB=90∘\boxed{\angle AOB = 90^\circ}

More questions in Exercise 10.2

Q1

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

  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
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) 60∘60^\circ (B) 70∘70^\circ (C) 80∘80^\circ (D) 90∘90^\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 80∘80^\circ, then ∠POA\angle \text{POA} is equal to (A) 50∘50^\circ (B) 60∘60^\circ (C) 70∘70^\circ (D) 80∘80^\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 A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Q7

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.

Q8

A quadrilateral ABCD 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, XY and X′Y′\text{X}'\text{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′\text{X}'\text{Y}' at B. Prove that ∠AOB=90∘\angle\text{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 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 Fig. 10.14). Find the sides AB and AC.

Q13

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

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