Board Previous Year Questions (Board PYQs)

Board Previous Year Questions

These are the question types the Boards ask on Circles. Work each one; the solution follows immediately.

PYQ 1 (1 mark): From a point $Q$ the length of the tangent to a circle is 24 cm and the distance of $Q$ from the centre is 25 cm. The radius is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm. [CBSE]

Solution: $r=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm. Answer: (A).

PYQ 2 (1 mark): If $TP,TQ$ are two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is: (A) 60° (B) 70° (C) 80° (D) 90°. [CBSE]

Solution: $\angle PTQ = 180^\circ-110^\circ = 70^\circ$. Answer: (B).

PYQ 3 (1 mark): If tangents $PA,PB$ from $P$ to a circle centre $O$ are inclined at $80^\circ$, then $\angle POA$ equals: (A) 50° (B) 60° (C) 70° (D) 80°. [CBSE]

Solution: $\angle AOB=100^\circ$; $OP$ bisects it, $\angle POA=50^\circ$. Answer: (A).

PYQ 4 (1 mark): The number of tangents that can be drawn to a circle from a point on the circle is: (A) 0 (B) 1 (C) 2 (D) infinite. [CBSE]

Solution: Exactly one. Answer: (B).

PYQ 5 (1 mark): A tangent at a point of a circle of radius 5 cm meets a line through the centre $O$ at $Q$ so that $OQ = 12$ cm. Length $PQ$ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D) $\sqrt{119}$ cm. [CBSE]

Solution: $PQ=\sqrt{12^2-5^2}=\sqrt{119}$ cm. Answer: (D).

PYQ 6 (2 marks): The length of a tangent from a point $A$ at distance 5 cm from the centre of a circle is 4 cm. Find the radius. [CBSE]

Solution: $r=\sqrt{5^2-4^2}=\sqrt{9}=3$ cm. Answer: 3 cm.

PYQ 7 (2 marks): Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE]

Solution: $2\sqrt{5^2-3^2}=2\sqrt{16}=8$ cm. Answer: 8 cm.

PYQ 8 (2 marks): Prove that the tangents drawn at the ends of a diameter of a circle are parallel. [CBSE]

Solution: Each tangent is perpendicular to the radius at its end; both radii lie along the diameter, so both tangents are perpendicular to the same line and hence parallel. $\blacksquare$

PYQ 9 (2 marks): Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. [CBSE]

Solution: The radius to the point of contact is perpendicular to the tangent (Theorem 10.1); through the point of contact there is only one perpendicular to the tangent, so it is the radius line, which passes through the centre. $\blacksquare$

PYQ 10 (3 marks): A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$. [CBSE]

Solution: Using equal tangents from each vertex ($AP=AS$, $BP=BQ$, $CR=CQ$, $DR=DS$), add: $AB+CD = (AP+BP)+(CR+DR) = (AS+BQ)+(CQ+DS) = (AS+DS)+(BQ+CQ) = AD+BC$. $\blacksquare$

PYQ 11 (3 marks): Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\,\angle OPQ$. [CBSE]

Solution: Let $\angle PTQ=\theta$. Since $TP=TQ$, $\angle TPQ=90^\circ-\tfrac{\theta}{2}$. As $\angle OPT=90^\circ$, $\angle OPQ = 90^\circ-(90^\circ-\tfrac{\theta}{2})=\tfrac{\theta}{2}$, so $\angle PTQ = 2\,\angle OPQ$. $\blacksquare$

PYQ 12 (3 marks): Prove that the lengths of tangents drawn from an external point to a circle are equal. [CBSE]

Solution: With tangents $PQ,PR$ and centre $O$: $\angle OQP=\angle ORP=90^\circ$, $OQ=OR$ (radii), $OP$ common $\Rightarrow \triangle OQP\cong\triangle ORP$ (RHS) $\Rightarrow PQ=PR$ (CPCT). $\blacksquare$

PYQ 13 (3 marks): Prove that the parallelogram circumscribing a circle is a rhombus. [CBSE]

Solution: For the circumscribing quadrilateral $AB+CD=AD+BC$; with $AB=CD$, $AD=BC$ this gives $AB=BC$, so all sides are equal — a rhombus. $\blacksquare$

PYQ 14 (3 marks): A triangle $ABC$ is drawn to circumscribe a circle of radius 4 cm such that $BD = 8$ cm and $DC = 6$ cm ($D$ = contact on $BC$). Find $AB$ and $AC$. [CBSE]

Solution: With $AF=AE=x$: $s=14+x$, Area $=4(14+x)=\sqrt{(14+x)\,x\cdot8\cdot6}$; squaring gives $x=7$. So $AB=15$ cm, $AC=13$ cm. Answer: $AB=15$ cm, $AC=13$ cm.

PYQ 15 (3 marks): 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. [CBSE]

Solution: In quadrilateral $OQPR$, $\angle OQP=\angle ORP=90^\circ$, so $\angle QPR+\angle QOR = 360^\circ-180^\circ = 180^\circ$ — they are supplementary. $\blacksquare$

PYQ 16 (3 marks): In the figure, $XY$ and $X'Y'$ are two parallel tangents to a circle centre $O$ and another tangent $AB$ with point of contact $C$ meets them at $A$ and $B$. Prove $\angle AOB = 90^\circ$. [CBSE]

Solution: $OA$ and $OB$ bisect the co-interior angles $\angle XAB$ and $\angle X'BA$ whose sum is $180^\circ$; so $\angle OAB+\angle OBA=90^\circ$ and $\angle AOB=90^\circ$. $\blacksquare$

PYQ 17 (2 marks): From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle. If $PA = 12$ cm, find $PB$. [CBSE]

Solution: Equal tangents: $PB = PA = 12$ cm.

PYQ 18 (4 marks): Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. [CBSE]

Solution: Joining the centre to the four contact points creates pairs of congruent triangles at each vertex (equal tangents + common radius). Summing the eight angles at $O$ to $360^\circ$ and pairing them shows $\angle AOB+\angle COD=180^\circ$ and $\angle BOC+\angle DOA=180^\circ$. $\blacksquare$

PYQ 19 (2 marks): In two concentric circles of radii 13 cm and 5 cm, find the length of the chord of the larger circle that touches the smaller. [CBSE]

Solution: $2\sqrt{13^2-5^2}=2\sqrt{144}=24$ cm.

PYQ 20 (1 mark): The length of the tangent from a point 13 cm away from the centre of a circle of radius 5 cm is: (A) 8 (B) 12 (C) 18 (D) $\sqrt{194}$ cm. [CBSE]

Solution: $\sqrt{13^2-5^2}=12$ cm. Answer: (B).