Circles | Exercise 10.2

Question 11

Prove that the parallelogram circumscribing a circle is a rhombus.

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Solution
Understand the Question
  • A parallelogram has equal opposite sides: AB=CDAB = CD and BC=ADBC = AD.
  • A rhombus is a parallelogram where all four sides are equal: AB=BC=CD=DAAB = BC = CD = DA.
  • When a quadrilateral circumscribes a circle, each side acts as a tangent. The lengths of tangents drawn from an external point to a circle are equal.
  • Summing the tangent lengths from all four vertices shows that the sum of opposite sides is equal (AB+CD=AD+BCAB + CD = AD + BC), which proves adjacent sides are equal (AB=BCAB = BC), confirming the figure is a rhombus.

Step 1 · Identify properties of the parallelogram

Let ABCDABCD be a parallelogram circumscribing a circle touching the sides ABAB, BCBC, CDCD, and DADA at points PP, QQ, RR, and SS respectively.Diagram 1

Since opposite sides of a parallelogram are equal:

AB=CD…(1)BC=AD…(2)\begin{aligned} AB &= CD \quad \dots (1) \\[0.6em] BC &= AD \quad \dots (2) \end{aligned}

Step 2 · Apply tangent theorem from external vertices

The lengths of tangents drawn from an external point to a circle are equal.

From vertex AA: AP=AS…(3)AP = AS \quad \dots (3)

From vertex BB: BP=BQ…(4)BP = BQ \quad \dots (4)

From vertex CC: CR=CQ…(5)CR = CQ \quad \dots (5)

From vertex DD: DR=DS…(6)DR = DS \quad \dots (6)

Step 3 · Sum the tangent equations and simplify

Adding equations (3)(3), (4)(4), (5)(5), and (6)(6): AP+BP+CR+DR=AS+BQ+CQ+DSAP + BP + CR + DR = AS + BQ + CQ + DS

Grouping adjacent segments to form the complete sides: (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB+CD=AD+BCAB + CD = AD + BC

Substitute CD=ABCD = AB from (1)(1) and AD=BCAD = BC from (2)(2):

AB+AB=BC+BC2AB=2BCAB=BC…(7)\begin{aligned} AB + AB &= BC + BC \\[0.6em] 2AB &= 2BC \\[0.6em] AB &= BC \quad \dots (7) \end{aligned}

Step 4 · Conclude that all four sides are equal

From equations (1)(1), (2)(2), and (7)(7): AB=BC=CD=DAAB = BC = CD = DA

Since all four sides of parallelogram ABCDABCD are equal, ABCDABCD is a rhombus.

Answer

Hence proved. The parallelogram ABCDABCD circumscribing a circle is a rhombus (AB=BC=CD=DAAB = BC = CD = DA).

Common Mistakes
  • Mismatched Tangent Grouping: When writing tangent equations, ensure segments belonging to the same side (like APAP and BPBP) are kept on the same side of the equation so they add up directly to ABAB.
  • Circular Logic: Assuming adjacent sides are equal at the start instead of using the tangent equality property to prove AB=BCAB = BC.

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) 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 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 X′Y′X'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 X′Y′X'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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