CBSE Class 10 Mathematics Circles Assignment Set 11

Board Level Exercise: Very Short Answer Questions (1 Mark Each)

Teacher's Note:
1. Use the angle in a semicircle.
2. The tangent is perpendicular to the diameter at A.

Teacher's Note:
1. Each tangent pair makes two equal angles at O.
2. The four angles at O add up to 360°.

Question 3. If the radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of one circle which is a tangent to the other circle.
Answer:
Half the chord = \( \sqrt{5^2 - 4^2} = 3 \) cm, because the radius of the smaller circle drawn to the point of contact is perpendicular to the chord.
The chord = 2 × 3 = 6 cm.
In simple words: Half the chord is the third side of a 3-4-5 right triangle.

Teacher's Note:
1. A radius at the point of contact is perpendicular to the tangent.
2. The perpendicular from the centre bisects the chord.

Teacher's Note:
1. Use the right angle between tangent and radius.
2. Then use the isosceles triangle OPQ.

Question 5. If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then find the length of each tangent.
Answer:
Let the tangents be PA and PB with ∠APB = 60°. The line PO bisects the angle, so ∠OPA = 30°, and OA ⊥ PA.
In the right triangle OAP, \( \tan 30^\circ = \dfrac{OA}{PA} = \dfrac{3}{PA} \), so \( PA = 3\sqrt{3} \) cm.
Each tangent is \( 3\sqrt{3} \) cm.
In simple words: The centre lies on the bisector of the angle between the tangents.

Teacher's Note:
1. Join the centre to the external point and to the point of contact.
2. Use tan 30° = 1/√3.

Teacher's Note:
1. The radius at the point of contact is perpendicular to the tangent.
2. Use cos 30° = √3/2.

Teacher's Note:
1. A chord parallel to the tangent has its mid-arc point at Q.
2. The base angles of the triangle are equal.

Short Answer Questions (2 Marks Each)

Question 8. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Answer:
Let the lines l and m meet at X and let a circle with centre O touch them at M and N. Join OM, ON and OX.
A tangent is perpendicular to the radius at the point of contact: ∠OMX = ∠ONX = 90°, and OM = ON (radii).
In the right triangles OMX and ONX: OM = ON and OX is common, so △OMX ≅ △ONX (R.H.S.). Hence ∠OXM = ∠OXN.
So OX bisects the angle between the lines: the centre lies on the angle bisector.
In simple words: Equal radii make two congruent right triangles, which gives equal angles at X.

Teacher's Note:
1. Draw the radii to the two points of contact.
2. Use the R.H.S. congruence rule.

Question 9. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.
Answer:
The line BO bisects ∠DBC, so ∠OBC = 60°. Also OC ⊥ BC, so △OCB is right-angled at C.
\( \cos 60^\circ = \dfrac{BC}{BO} \), so \( BC = \dfrac{BO}{2} \), that is BO = 2BC.
BD = BC (tangents from an external point), so BC + BD = 2BC = BO. Hence proved.
In simple words: The centre lies on the bisector of the angle between the tangents, giving a 30-60-90 triangle.

Teacher's Note:
1. Join OC and OB.
2. Use cos 60° = 1/2.

Question 10. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.
Answer:
Since BC ⊥ AB at the end B of the diameter, BC is a tangent to the circle at B. Let the tangent at P meet BC at Q. Then QB = QP (tangents from Q), so ∠QBP = ∠QPB.
∠APB = 90° (angle in a semicircle), so ∠BPC = 90° and in the right triangle BPC, ∠QBP + ∠QCP = 90° and ∠QPB + ∠QPC = 90°.
Because ∠QBP = ∠QPB, we get ∠QPC = ∠QCP, so QP = QC.
Hence QB = QP = QC: the tangent at P bisects BC.
In simple words: Q is equidistant from B, P and C, so it is the midpoint of BC.

Teacher's Note:
1. BC touches the circle at B.
2. Use two pairs of tangent lengths from Q.

Teacher's Note:
1. Tangent lengths from P are equal.
2. Use alternate angles and the alternate segment theorem.

Question 12. Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.
Answer:
Let CD be a chord parallel to the tangent at A. The tangent at A is perpendicular to the radius OA, that is, to the diameter AB. Since CD is parallel to the tangent, CD ⊥ AB.
The perpendicular drawn from the centre of a circle to a chord bisects the chord. OM ⊥ CD, where M is the point where AB meets CD, so CM = MD.
Hence AB bisects every such chord.
In simple words: The diameter through the point of contact is perpendicular to the tangent, so it is also perpendicular to every parallel chord.

Teacher's Note:
1. A tangent is perpendicular to the radius at the contact point.
2. A perpendicular from the centre bisects a chord.

Long Answer Questions (3 Marks Each)

Question 13. Two circles with centres O and O' of radii 3 cm and 4 cm, respectively, intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.
Answer:
OP is a tangent to the circle with centre O' and O'P is a tangent to the circle with centre O, so the radii OP and O'P are perpendicular: ∠OPO' = 90°. In the right triangle OPO', OO' = \( \sqrt{3^2 + 4^2} \) = 5 cm.
The common chord PQ is bisected at right angles by OO' at a point M, and the area of △OPO' gives \( \frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 5 \times PM \), so PM = 2.4 cm.
PQ = 2 × PM = 4.8 cm.
In simple words: The common chord is twice the altitude from the right angle in the 3-4-5 triangle.

Teacher's Note:
1. The line of centres is the perpendicular bisector of the common chord.
2. Compare two expressions for the area of the right triangle.

Question 14. If a hexagon ABCDEF circumscribes a circle, prove that AB + CD + EF = BC + DE + FA.
Answer:
Let the tangent lengths from A, B, C, D, E, F be a, b, c, d, e, f (tangents from an external point are equal).
AB = a + b, BC = b + c, CD = c + d, DE = d + e, EF = e + f, FA = f + a.
AB + CD + EF = (a + b) + (c + d) + (e + f) = a + b + c + d + e + f and BC + DE + FA = (b + c) + (d + e) + (f + a) = a + b + c + d + e + f.
Hence AB + CD + EF = BC + DE + FA.
In simple words: Each side splits into two tangent lengths, and both sums use every tangent length once.

Teacher's Note:
1. Name the tangent length from each vertex.
2. Add alternate sides.

Teacher's Note:
1. Use the angle in a semicircle.
2. Use the right angle between tangent and diameter.

Very Long Answer Questions (4 Marks Each)

Question 16. A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the △ABC.
Answer:
Tangents from an external point are equal: AP = AQ, BP = BR and CQ = CR.
In the right triangle OPA, \( AP = \sqrt{13^2 - 5^2} = 12 \) cm, so AQ = 12 cm.
Perimeter of △ABC = AB + BC + CA = AB + BR + RC + CA = AB + BP + CQ + CA = AP + AQ = 12 + 12 = 24 cm.
In simple words: The perimeter of the triangle formed by the tangent BC and the two tangents equals AP + AQ.

Teacher's Note:
1. Replace BR by BP and CR by CQ.
2. Find AP with Pythagoras theorem.

Question 17. If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.
Answer:
Let AD be the altitude from A to BC. It passes through the centre, and meeting the circle again at E gives the diameter AE = 18 cm.
In the right triangle ABE (∠ABE = 90°, angle in a semicircle), BD ⊥ AE, so \( AB^2 = AD \times AE \): \( 36 = AD \times 18 \), so AD = 2 cm.
\( BD = \sqrt{AB^2 - AD^2} = \sqrt{36 - 4} = 4\sqrt{2} \) cm, so BC = \( 8\sqrt{2} \) cm.
Area = \( \dfrac{1}{2} \times BC \times AD = \dfrac{1}{2} \times 8\sqrt{2} \times 2 = \) \( 8\sqrt{2} \) cm².
In simple words: The altitude of the isosceles triangle is a part of the diameter of the circumscribed circle.

Teacher's Note:
1. In a right triangle, the square of a leg equals the hypotenuse times its projection.
2. Use the diameter through A.

Teacher's Note:
1. Use the perpendicular from the centre to the chord.
2. Use Pythagoras theorem in three right triangles.