RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
In Class 9 you treated a circle as a set of points; here you study what happens when a straight line meets that circle. Almost every question in this chapter is unlocked by just two facts — the tangent is perpendicular to the radius at the point of contact, and the two tangents from an outside point have equal length. Train yourself to redraw the figure, mark the right angle at each point of contact, and then hunt for a right triangle or a pair of congruent triangles.
PQ is a tangent at the point P to a circle of radius 8 cm with centre O, and Q lies on a line through O with OQ = 17 cm. The length of PQ is:
Answer
The length of PQ is 15 cm. Since PQ is a tangent at P, the radius OP is perpendicular to PQ, so ΔOPQ is right-angled at P. By Pythagoras Theorem, OQ² = OP² + PQ². 17² = 8² + PQ² 289 = 64 + PQ², so PQ² = 225. Therefore PQ = 15 cm.
Two circles with the same centre have radii 5 cm and 13 cm. The length of the chord of the larger circle that just touches the smaller circle is:
Answer
The chord is 24 cm long. Let the common centre be O and let the chord AB of the larger circle touch the smaller circle at M. Since AB is a tangent to the smaller circle at M, OM ⊥ AB and OM = 5 cm. A perpendicular from the centre bisects the chord, so AM = MB. In right triangle OMA, OA = 13 cm, so AM² = 13² - 5² = 169 - 25 = 144, giving AM = 12 cm. Hence AB = 2 × 12 = 24 cm.
The number of tangents that can be drawn to a circle from a point lying in its interior is:
Answer
The correct answer is 0. Every straight line through an interior point must cross the circle, and it crosses it at two distinct points. Such a line is a secant, not a tangent, because a tangent meets a circle at exactly one point. So no tangent at all can be drawn from an interior point. (From a point on the circle there is exactly one, and from an external point exactly two.)
PA and PB are two tangents drawn from an external point P to a circle with centre O, touching it at A and B. If ∠APB = 70°, then ∠AOB equals:
Answer
∠AOB = 110°. OA ⊥ PA and OB ⊥ PB, so ∠OAP = ∠OBP = 90°. In quadrilateral OAPB the four angles add up to 360°: ∠AOB + ∠OAP + ∠APB + ∠OBP = 360° ∠AOB + 90° + 70° + 90° = 360° ∠AOB = 360° - 250° = 110°. This matches the general rule ∠AOB + ∠APB = 180°.
Assertion (A): The two tangents drawn to a circle from a single external point always have equal lengths. Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer
Both A and R are true and R is the correct explanation of A. Let PA and PB be the tangents from P to a circle with centre O, touching at A and B. By Reason R, ∠OAP = ∠OBP = 90°, so ΔOAP and ΔOBP are right triangles. They share the hypotenuse OP, and OA = OB since both are radii. By the RHS congruence rule, ΔOAP ≅ ΔOBP, so the corresponding sides give PA = PB. The right angle supplied by R is exactly what makes this proof work, so R correctly explains A.
A tangent is drawn from an external point B to a circle of radius 7 cm with centre O, touching the circle at A. If OB = 25 cm, find the length of the tangent AB.
Answer
Since AB is a tangent at A, OA ⊥ AB, so ΔOAB is right-angled at A. By Pythagoras Theorem, OB² = OA² + AB². 25² = 7² + AB² 625 = 49 + AB², so AB² = 576. Therefore AB = 24 cm.
A quadrilateral ABCD is drawn so that all four of its sides touch a circle. If AB = 9 cm, BC = 11 cm and CD = 8 cm, find the length of AD. Justify the result you use.
Answer
Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively. Tangents from the same external point are equal, so: AP = AS, BP = BQ, CQ = CR and DR = DS. Adding the sides AB and CD: AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS). Adding the sides BC and AD: BC + AD = (BQ + QC) + (AS + SD) = (BQ + CQ) + (AS + DS). The two totals contain exactly the same four lengths, so AB + CD = BC + AD. Substituting the given values: 9 + 8 = 11 + AD. 17 = 11 + AD, so AD = 6 cm.
From an external point T, two tangents TP and TQ are drawn to a circle of radius 6 cm with centre O. If ∠PTQ = 60°, find the length of the tangent TP.
Answer
OT bisects ∠PTQ, so ∠OTP = 30°, and OP ⊥ TP gives a right angle at P. In right triangle OTP, tan 30° = OP/TP = 6/TP. Since tan 30° = 1/√3, we get 1/√3 = 6/TP. Therefore TP = 6√3 cm ≈ 10.39 cm.
Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this result, solve: tangents PA and PB are drawn from an external point P to a circle, and a third tangent touching the circle at C cuts PA at D and PB at E. If PA = 14 cm, find the perimeter of ΔPDE.
Answer
Proof: Let the circle have centre O and let PA and PB be tangents from the external point P, touching at A and B. Step 1 — Join OA, OB and OP. Step 2 — A tangent is perpendicular to the radius at the point of contact, so ∠OAP = ∠OBP = 90°. Step 3 — In ΔOAP and ΔOBP: OA = OB (radii of the same circle), OP = OP (common hypotenuse), ∠OAP = ∠OBP = 90°. Step 4 — By the RHS congruence rule, ΔOAP ≅ ΔOBP. Step 5 — Corresponding parts of congruent triangles are equal, so PA = PB. Hence proved. Application: D lies on PA and the third tangent touches at C, so DA and DC are tangents from D, giving DA = DC. Similarly E lies on PB, so EB = EC. Perimeter of ΔPDE = PD + DE + EP = PD + (DC + CE) + EP = PD + DA + EB + EP (replacing DC by DA and CE by EB) = (PD + DA) + (EP + EB) = PA + PB. Since PA = PB = 14 cm, the perimeter = 14 + 14 = 28 cm.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Then use tangent properties to solve: a circle is inscribed in ΔABC, touching BC at D, CA at E and AB at F. If BD = 5 cm, CD = 7 cm and AF = 4 cm, find the perimeter of ΔABC.
Answer
Proof: Let the circle have centre O and let the line ℓ be the tangent touching the circle at the point P. We must show OP ⊥ ℓ. Step 1 — Take any point Q on ℓ other than P. Step 2 — Since ℓ touches the circle only at P, every such Q lies outside the circle, so OQ > OP. Step 3 — This is true for every point Q on ℓ, so OP is the shortest of all the segments joining O to points of ℓ. Step 4 — The shortest segment from a point to a line is the perpendicular segment. Step 5 — Therefore OP ⊥ ℓ, that is, the tangent is perpendicular to the radius at the point of contact. Hence proved. Application: Tangents from the same external point are equal, so: BF = BD = 5 cm (from B), CE = CD = 7 cm (from C), AE = AF = 4 cm (from A). Now compute the sides: AB = AF + FB = 4 + 5 = 9 cm. AC = AE + EC = 4 + 7 = 11 cm. BC = BD + DC = 5 + 7 = 12 cm. Perimeter of ΔABC = 9 + 11 + 12 = 32 cm.
Read the following and answer the questions that follow: A circular ornamental fountain of radius 9 m has its centre at O. A lamp post stands at a point P on the flat ground outside the fountain, with OP = 41 m. Two straight decorative wires are stretched from the top of the lamp post's base at P so that each just touches the rim of the fountain, at the points A and B. (a) Find the length of the wire PA. (b) Write the length of PB and state the property used. (c) Find the area of the quadrilateral OAPB. (d) If the two wires make an angle ∠APB = 60° with each other, find ∠AOB.
Answer
(a) PA touches the circle at A, so OA ⊥ PA and ΔOAP is right-angled at A. OP² = OA² + PA² 41² = 9² + PA² 1681 = 81 + PA², so PA² = 1600 and PA = 40 m. (b) PB = 40 m as well, because the lengths of the two tangents drawn from an external point to a circle are equal. (c) Quadrilateral OAPB is made of the two right triangles OAP and OBP. Area of ΔOAP = (1/2) × OA × PA = (1/2) × 9 × 40 = 180 m². The two triangles are congruent, so area of ΔOBP = 180 m². Area of OAPB = 180 + 180 = 360 m². (d) In quadrilateral OAPB, ∠OAP = ∠OBP = 90°, and the angle sum is 360°. ∠AOB + 90° + 60° + 90° = 360° ∠AOB = 360° - 240° = 120°.
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