Mathematics: Circles (Tangents & Secants)
Class X · Time: 3 Hours · Maximum Marks: 80
General Instructions
- All questions are compulsory. The paper has 31 questions in 5 sections.
- Section A: Q1 to 10, MCQ, 1 mark each. Section B: Q11 to 16, 2 marks each. Section C: Q17 to 22, 3 marks each. Section D: Q23 to 27, 4 marks each. Section E: Q28 to 31, 5 marks each.
- Draw neat figures wherever needed. Show all steps; marks are given for steps.
Section A: Multiple Choice (10 × 1 = 10)
11. A tangent to a circle intersects it in how many points?
(a) 0(b) 1(c) 2(d) infinite
12. A line intersecting a circle in two points is called a:
(a) tangent(b) chord(c) secant(d) diameter
13. A circle can have at most how many tangents parallel to a given secant?
(a) 1(b) 2(c) 3(d) 4
14. A tangent $PQ$ at $P$ to a circle of radius $5$ cm meets a line through the centre $O$ at $Q$, with $OQ=13$ cm. Then $PQ=$
(a) 12 cm(b) 8 cm(c) 18 cm(d) $\sqrt{194}$ cm
15. The angle between a tangent and the radius at the point of contact is:
(a) $45^\circ$(b) $60^\circ$(c) $90^\circ$(d) $180^\circ$
16. The number of tangents from a point inside a circle is:
(a) 0(b) 1(c) 2(d) 4
17. Tangents $PA$ and $PB$ are drawn from $P$ to a circle. If $PA=7$ cm, then $PB=$
(a) 3.5 cm(b) 7 cm(c) 14 cm(d) 49 cm
18. A line is at a distance $6$ cm from the centre of a circle of radius $6$ cm. The line is:
(a) a secant(b) a tangent(c) non-intersecting(d) a chord
19. Tangents $TP$, $TQ$ touch a circle with centre $O$ and $\angle POQ=100^\circ$. Then $\angle PTQ=$
(a) $50^\circ$(b) $60^\circ$(c) $80^\circ$(d) $100^\circ$
110. Tangents $PA$, $PB$ from $P$ are inclined at $60^\circ$. Then $\angle POA=$
(a) $30^\circ$(b) $60^\circ$(c) $90^\circ$(d) $120^\circ$
Section B: Short Answer I (6 × 2 = 12)
211. Define a tangent to a circle and its point of contact.
212. The tangent from a point $Q$ to a circle of radius $7$ cm is $24$ cm long. Find the distance of $Q$ from the centre.
213. Why can no tangent be drawn to a circle from a point inside it?
214. Tangents $TP$ and $TQ$ are drawn from $T$ and $\angle PTQ=40^\circ$. Find $\angle TPQ$.
215. Two concentric circles have radii $10$ cm and $6$ cm. Find the length of the chord of the larger circle that touches the smaller circle.
216. A quadrilateral $ABCD$ circumscribes a circle, with $AB=9$ cm, $BC=7$ cm, $CD=6$ cm. Find $AD$.
Section C: Short Answer II (6 × 3 = 18)
317. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
318. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
319. The length of a tangent from a point $A$ at distance $17$ cm from the centre $O$ is $15$ cm. Find the radius of the circle and the area of the triangle $OPA$, where $P$ is the point of contact.
320. Tangents $TP$ and $TQ$ are drawn from $T$ to a circle with centre $O$, and $\angle POQ=130^\circ$. Find $\angle PTQ$ and $\angle OPQ$.
321. Prove that a parallelogram circumscribing a circle is a rhombus.
322. $PQ$ is a chord of length $24$ cm of a circle of radius $13$ cm. The tangents at $P$ and $Q$ meet at $T$. Find $TP$.
Section D: Long Answer I (5 × 4 = 20)
423. Prove that the lengths of tangents drawn from an external point to a circle are equal.
424. 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.
425. A quadrilateral $ABCD$ circumscribes a circle. Prove that $AB+CD=AD+BC$.
426. In a right triangle $ABC$, $\angle B=90^\circ$, $AB=12$ cm and $BC=16$ cm. A circle is drawn inside the triangle touching all three sides. Find its radius.
427. Two tangents $PA$ and $PB$ are drawn from $P$ to a circle with centre $O$ and radius $6$ cm, where $OP=10$ cm. Find (i) the length of each tangent, (ii) the area of quadrilateral $OAPB$, (iii) the length of chord $AB$.
Section E: Long Answer II (4 × 5 = 20)
528. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
529. Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ=2\angle OPQ$. Hence find $\angle TPQ$ if $\angle PTQ=70^\circ$.
530. $XY$ and $X'Y'$ are parallel tangents to a circle with centre $O$, touching it at $P$ and $Q$. Another tangent $AB$ touches the circle at $C$ and meets $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB=90^\circ$.
531. Two concentric circles have radii $13$ cm and $5$ cm. A chord $AB$ of the larger circle touches the smaller circle at $P$. Prove that $AP=BP$ and find the length of $AB$.
📝 Answer Key and Marking Scheme
Section A (1 mark each)
1. (b) 1 2. (c) secant 3. (b) 2 5. (c) $90^\circ$ 6. (a) 0 7. (b) 7 cm 8. (b) tangent ($d=r$)
4. (a) 12 cm. $OP\perp PQ$ (Theorem 10.1), so $PQ^2=OQ^2-OP^2=169-25=144$, $PQ=12$.
9. (c) $80^\circ$. In quadrilateral $OPTQ$: $90^\circ+90^\circ+100^\circ+\angle PTQ=360^\circ$, so $\angle PTQ=80^\circ$.
10. (b) $60^\circ$. $OP$ bisects the angle: $\angle APO=30^\circ$. In $\triangle OAP$: $\angle POA=180^\circ-90^\circ-30^\circ=60^\circ$.
Section B (2 marks each)
11. A tangent is a line that meets the circle at exactly one point (1). That common point is the point of contact (1).
12. $OP\perp PQ$, so $OQ^2=OP^2+PQ^2=7^2+24^2=49+576=625$ (1). $OQ=\mathbf{25}$ cm (1).
13. Every line through an interior point enters and leaves the circle, so it meets the circle at two points (1). That makes it a secant; a tangent needs exactly one common point (1).
14. $TP=TQ$ (Theorem 10.2), so $\triangle TPQ$ is isosceles (1). $\angle TPQ=\tfrac{180^\circ-40^\circ}{2}=\mathbf{70^\circ}$ (1).
15. The chord $AB$ touches the smaller circle at $P$, so $OP\perp AB$, $OP=6$. In $\triangle OPA$: $AP^2=10^2-6^2=64$, $AP=8$ (1). $OP$ bisects $AB$, so $AB=2\times8=\mathbf{16}$ cm (1).
16. For a circumscribed quadrilateral $AB+CD=AD+BC$ (1). $9+6=AD+7$, so $AD=\mathbf{8}$ cm (1).
Section C (3 marks each)
17. Let $PQ$ be a diameter, with tangents $XY$ at $P$ and $X'Y'$ at $Q$ (1). The diameter passes through $O$, so $PQ$ lies along the radius at both points. By Theorem 10.1, $PQ\perp XY$ and $PQ\perp X'Y'$ (1). So $\angle XPQ=\angle PQY'=90^\circ$ (alternate angles are equal), hence $XY\parallel X'Y'$ (1).
18. Let $XY$ touch the circle at $P$ with centre $O$ (1). By Theorem 10.1, $OP\perp XY$ (1). Only one line can be perpendicular to $XY$ at the point $P$, so the perpendicular at $P$ is the line $OP$, which passes through the centre (1). ∎
19. $\angle OPA=90^\circ$, so $OP^2=OA^2-AP^2=289-225=64$, radius $OP=\mathbf{8}$ cm (2). Area $=\tfrac12\times OP\times AP=\tfrac12\times8\times15=\mathbf{60}\text{ cm}^2$ (1).
20. $OPTQ$ has $\angle OPT=\angle OQT=90^\circ$, so $\angle PTQ=360^\circ-90^\circ-90^\circ-130^\circ=\mathbf{50^\circ}$ (1). $TP=TQ$, so $\angle TPQ=\tfrac{180^\circ-50^\circ}{2}=65^\circ$ (1). $\angle OPQ=\angle OPT-\angle TPQ=90^\circ-65^\circ=\mathbf{25^\circ}$ (1). Check: $\angle PTQ=2\angle OPQ$ ✓
21. Let parallelogram $ABCD$ circumscribe a circle (1). Then $AB+CD=AD+BC$ (tangents from each vertex are equal). In a parallelogram $AB=CD$ and $AD=BC$ (1). So $2AB=2AD$, so $AB=AD$. Adjacent sides are equal, so $ABCD$ is a rhombus (1). ∎
22. $OT$ meets $PQ$ at $R$; $OT\perp PQ$ and $PR=12$ (1). $OR=\sqrt{13^2-12^2}=5$ (1). $\angle RPO=\angle PTR$, so right $\triangle TRP\sim\triangle PRO$ (AA): $\dfrac{TP}{PO}=\dfrac{RP}{RO}\Rightarrow\dfrac{TP}{13}=\dfrac{12}{5}$, so $TP=\dfrac{156}{5}=\mathbf{31.2}$ cm (1).
Section D (4 marks each)
23. Given: circle with centre $O$, external point $P$, tangents $PQ$, $PR$. To prove: $PQ=PR$ (1). Join $OP,OQ,OR$. By Theorem 10.1, $\angle OQP=\angle ORP=90^\circ$ (1). In right triangles $OQP$ and $ORP$: $OQ=OR$ (radii), $OP=OP$ (common) (1). So $\triangle OQP\cong\triangle ORP$ (RHS), hence $PQ=PR$ (CPCT). ∎ (1)
24. Let tangents $PA,PB$ touch the circle (centre $O$) at $A,B$ (1). $\angle OAP=\angle OBP=90^\circ$ by Theorem 10.1 (1). Angle sum of quadrilateral $OAPB$: $\angle AOB+\angle APB+90^\circ+90^\circ=360^\circ$ (1). So $\angle AOB+\angle APB=180^\circ$, i.e. they are supplementary. ∎ (1)
25. Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$ (1). Theorem 10.2: $AP=AS$, $BP=BQ$, $CR=CQ$, $DR=DS$ (1). $AB+CD=(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)$ (1) $=(AS+DS)+(BQ+CQ)=AD+BC$. ∎ (1)
26. $AC=\sqrt{12^2+16^2}=20$ (1). Let the circle (radius $r$) touch $AB,BC,CA$ at $F,D,E$. $OFBD$ has three right angles and $OF=OD=r$, so it is a square: $BF=BD=r$ (1). $AE=AF=12-r$ and $CE=CD=16-r$ (Theorem 10.2) (1). $AC=AE+CE$: $20=28-2r\Rightarrow r=\mathbf{4}$ cm (1). Check: Area $=96$, $s=24$, $r=96/24=4$ ✓
27. (i) $OA\perp PA$, so $PA^2=10^2-6^2=64$, $PA=PB=\mathbf8$ cm (1). (ii) Area $=2\times\tfrac12\times6\times8=\mathbf{48}\text{ cm}^2$ (1). (iii) $OP\perp AB$ at its midpoint $R$. The area of $OAPB$ is also $\tfrac12\times OP\times AB$ (1), so $\tfrac12\times10\times AB=48$, $AB=\mathbf{9.6}$ cm (1).
Section E (5 marks each)
28. Given: circle, centre $O$, tangent $XY$ at $P$. To prove: $OP\perp XY$ (1). Take any point $Q\ne P$ on $XY$ and join $OQ$ (1). $Q$ must lie outside the circle: if it were inside, $XY$ would be a secant, not a tangent (1). So $OQ>OP$ (1). This holds for every $Q\ne P$ on $XY$, so $OP$ is the shortest distance from $O$ to $XY$, and the shortest distance is the perpendicular. So $OP\perp XY$. ∎ (1)
29. Let $\angle PTQ=\theta$. $TP=TQ$, so $\angle TPQ=\tfrac12(180^\circ-\theta)=90^\circ-\tfrac{\theta}{2}$ (1). $\angle OPT=90^\circ$, so $\angle OPQ=90^\circ-\left(90^\circ-\tfrac{\theta}{2}\right)=\tfrac{\theta}{2}$ (1). Hence $\angle PTQ=2\angle OPQ$. ∎ (1). For $\theta=70^\circ$: $\angle OPQ=35^\circ$ (1), so $\angle TPQ=90^\circ-35^\circ=\mathbf{55^\circ}$ (1).
30. Tangents from $A$: $AP=AC$, so $OA$ bisects $\angle PAC$, i.e. $\angle OAB=\tfrac12\angle PAB$ (1). Tangents from $B$: $BQ=BC$, so $\angle OBA=\tfrac12\angle QBA$ (1). $XY\parallel X'Y'$ with transversal $AB$: $\angle PAB+\angle QBA=180^\circ$ (co-interior) (1). So $\angle OAB+\angle OBA=\tfrac12(180^\circ)=90^\circ$ (1). In $\triangle AOB$: $\angle AOB=180^\circ-90^\circ=\mathbf{90^\circ}$. ∎ (1)
31. $AB$ is a tangent to the smaller circle at $P$, so $OP\perp AB$ (Theorem 10.1) (1). $AB$ is a chord of the larger circle and the perpendicular from the centre bisects a chord, so $AP=BP$. ∎ (2). In right $\triangle OPA$: $AP^2=OA^2-OP^2=13^2-5^2=144$, $AP=12$ (1). $AB=2\times12=\mathbf{24}$ cm (1).
