CBSE Class 10 Mathematics Circles Assignment Set 12

Previous Year Problems on Circles

Teacher's Note:
1. Name the three tangent lengths x, y, z.
2. Solve the three equations together.

Teacher's Note:
1. A quadrilateral with four right angles and two adjacent equal sides is a square.

Teacher's Note:
1. In a right-angled corner, the radius equals the tangent length from that corner.

Question 4. In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that ∠DOE = 90°. [4 Marks]
Answer:
Since l ∥ m and OA ⊥ l, OB ⊥ m, the points A, O, B lie on one straight line (AB is a diameter), so ∠AOC + ∠COB = 180°.
Tangents DA and DC from D are equal and OA = OC, OD = OD, so △DAO ≅ △DCO (S.S.S.) and ∠AOD = ∠DOC = ½∠AOC. Similarly ∠COE = ∠EOB = ½∠COB.
∠DOE = ∠DOC + ∠COE = ½(∠AOC + ∠COB) = ½ × 180° = 90°.
In simple words: OD and OE bisect two angles that together make a straight angle.

Teacher's Note:
1. The centre lies on the bisector of the angle between two tangents.
2. Add the two half-angles.

Question 5. Two concentric circles are of radii 5 cm and 3 cm. The length of the chord of the larger circle (in cm) which touches the smaller circle is: [1 Mark]
(A) 4   (B) 5   (C) 8   (D) 10
Answer:
Ans. (C) 8
Half the chord = \( \sqrt{5^2 - 3^2} = 4 \) cm, so the chord is 8 cm.
In simple words: The radius of the smaller circle is perpendicular to the chord. | Use Pythagoras theorem.

Teacher's Note:
1. The chord is twice the leg of the 3-4-5 triangle.

Teacher's Note:
1. Add the tangent lengths from A and from B.

Question 7. Prove that a quadrilateral drawn to circumscribe a circle has the sums of the opposite sides equal. [4 Marks]
Answer:
Let ABCD circumscribe a circle touching AB, BC, CD, DA at P, Q, R, S. Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), that is AB + CD = AD + BC.
Hence the sums of opposite sides are equal.
In simple words: Each side splits into two tangent lengths, and each tangent length appears once on each side of the equation.

Teacher's Note:
1. Name the four points of contact.
2. Add the four tangent equalities.

Teacher's Note:
1. Replace AR by AP and BR by BQ.
2. Use XP = XQ.

Teacher's Note:
1. Use the tangent-chord angle.
2. A cyclic quadrilateral has opposite angles that add up to 180°.

Teacher's Note:
1. Join OP.
2. Use the equal tangent lengths from R.

Question 11. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. [2 Marks]
Answer:
Let M be the mid-point of the arc AB, so arc AM = arc MB and chord AM = chord MB. Also OA = OB.
Both O and M are equidistant from A and B, so the line OM is the perpendicular bisector of AB: OM ⊥ AB.
The tangent at M is perpendicular to OM. Two lines perpendicular to the same line OM are parallel, so the tangent at M is parallel to AB.
In simple words: The radius to the mid-point of the arc is perpendicular to the chord, and the tangent is perpendicular to that radius.

Teacher's Note:
1. Show that OM ⊥ AB.
2. Then use the tangent being perpendicular to OM.

Teacher's Note:
1. Use sin of the angle at P.
2. Triangle OTS is isosceles.

Teacher's Note:
1. Find AO' and AO in terms of r.
2. The ratio of corresponding sides is the required ratio.

Question 14. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O is 60°, then find the length of OP. [1 Mark]
Answer:
∠OPA = 30° (OP bisects the angle between the tangents) and OA ⊥ PA.
In the right triangle OAP, \( \sin 30^\circ = \dfrac{OA}{OP} = \dfrac{a}{OP} = \dfrac{1}{2} \), so OP = 2a.
In simple words: In a 30-60-90 triangle, the hypotenuse is twice the shortest side.

Teacher's Note:
1. The line OP bisects the angle between the tangents.
2. The radius is perpendicular to the tangent.