\displaystyle \textbf{Question 1: }\text{In the adjoining figure, }PA\text{ and }PB\text{ are tangents to the circle drawn from an external}
\displaystyle \text{point }P\text{. }CD\text{ is a third tangent touching the circle at }Q\text{. If }PB=10\text{ cm and}
\displaystyle CQ=2\text{ cm, what is the length }PC\text{?} \displaystyle \text{Answer:}
\displaystyle PA=PB=10\text{ cm}\qquad\text{[Tangents from }P\text{]}
\displaystyle CA=CQ=2\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle PA=PC+CA
\displaystyle \therefore PC=PA-CA=10-2=8\text{ cm}.
\displaystyle \therefore \text{The length of }PC\text{ is }8\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Two tangents }TP\text{ and }TQ\text{ are drawn from an external point }T\text{ to a circle}
\displaystyle \text{with centre }O\text{ as shown in In the adjoining figure,. If they are inclined to each other at an angle of }100^\circ,
\displaystyle \text{then what is the value of }\angle POQ\text{?} \displaystyle \text{Answer:}
\displaystyle \text{The angle between two tangents from an external point and the angle subtended by the}
\displaystyle \text{points of contact at the centre are supplementary.}
\displaystyle \therefore \angle POQ+\angle PTQ=180^\circ
\displaystyle \angle POQ+100^\circ=180^\circ
\displaystyle \therefore \angle POQ=80^\circ.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }CP\text{ and }CQ\text{ are tangents to a circle with centre }O\text{. }ARB
\displaystyle \text{is another tangent touching the circle at }R\text{. If }CP=11\text{ cm and }BC=7\text{ cm,}
\displaystyle \text{then find the length of }BR\text{.}\hfill\text{[CBSE 2009]} \displaystyle \text{Answer:}
\displaystyle CP=CQ=11\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle BQ=CQ-BC=11-7=4\text{ cm}.
\displaystyle BR=BQ\qquad\text{[Tangents from }B\text{]}
\displaystyle \therefore BR=4\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The length of tangent from a point }A\text{ at a distance of }5\text{ cm from the centre}
\displaystyle \text{of the circle is }4\text{ cm. What is the radius of the circle?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }O\text{ be the centre and }T\text{ be the point of contact.}
\displaystyle OA=5\text{ cm},\qquad AT=4\text{ cm}
\displaystyle \text{Since the radius is perpendicular to the tangent at the point of contact, }OT\perp AT.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle OA^2=OT^2+AT^2
\displaystyle 5^2=OT^2+4^2
\displaystyle OT^2=25-16=9
\displaystyle \therefore OT=3\text{ cm}.
\displaystyle \therefore \text{The radius of the circle is }3\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the distance between two parallel tangents to a circle of radius }5\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{The distance between two parallel tangents to a circle is equal to its diameter.}
\displaystyle \therefore \text{Distance}=2r=2\times5=10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In Question No. 1, if }PB=10\text{ cm, what is the perimeter of }\triangle PCD\text{?}
\displaystyle \text{Answer:}
\displaystyle PA=PB=10\text{ cm}\qquad\text{[Tangents from }P\text{]}
\displaystyle CA=CQ=2\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle \therefore PC=PA-CA=10-2=8\text{ cm}.
\displaystyle DB=DQ\qquad\text{[Tangents from }D\text{]}
\displaystyle PB=PD+DB=PD+DQ=10\text{ cm}.
\displaystyle CD=CQ+QD.
\displaystyle \text{Perimeter of }\triangle PCD=PC+PD+CD
\displaystyle =PC+(PD+DQ)+CQ
\displaystyle =8+10+2=20\text{ cm}.
\displaystyle \therefore \text{The perimeter of }\triangle PCD\text{ is }20\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{What is the distance between two parallel tangents of a circle of radius }4\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{The distance between two parallel tangents to a circle is equal to its diameter.}
\displaystyle \therefore \text{Distance}=2r=2\times4=8\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }\triangle ABC\text{ is circumscribing a circle. Find the length of }BC\text{.}
\displaystyle \hfill\text{[CBSE 2009, 20]} \displaystyle \text{Answer:}
\displaystyle AR=AQ=4\text{ cm}\qquad\text{[Tangents from }A\text{]}
\displaystyle BR=BP=3\text{ cm}\qquad\text{[Tangents from }B\text{]}
\displaystyle AC=11\text{ cm}.
\displaystyle CQ=AC-AQ=11-4=7\text{ cm}.
\displaystyle CP=CQ=7\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle BC=BP+PC=3+7=10\text{ cm}.
\displaystyle \therefore \text{The length of }BC\text{ is }10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }CP\text{ and }CQ\text{ are tangents from an external point }C\text{ to a circle}
\displaystyle \text{with centre }O\text{. }AB\text{ is another tangent which touches the circle at }R\text{. If }CP=11\text{ cm}
\displaystyle \text{and }BR=4\text{ cm, find the length of }BC\text{.}\hfill\text{[CBSE 2010]} \displaystyle \text{Answer:}
\displaystyle CP=CQ=11\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle BR=BQ=4\text{ cm}\qquad\text{[Tangents from }B\text{]}
\displaystyle CQ=CB+BQ
\displaystyle \therefore BC=CQ-BQ=11-4=7\text{ cm}.
\displaystyle \therefore \text{The length of }BC\text{ is }7\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two concentric circles are of radii }a\text{ and }b\ (a>b)\text{. Find the length of}
\displaystyle \text{the chord of the larger circle which touches the smaller circle.}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the chord of the larger circle touching the smaller circle at }P.
\displaystyle \text{Let }O\text{ be the common centre of the two circles.}
\displaystyle \text{Since }AB\text{ is tangent to the smaller circle at }P,\ OP\perp AB.
\displaystyle \text{Also, the perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore AP=PB.
\displaystyle OA=a\qquad\text{and}\qquad OP=b.
\displaystyle \text{In right triangle }OAP,
\displaystyle OA^2=OP^2+AP^2
\displaystyle a^2=b^2+AP^2
\displaystyle \therefore AP=\sqrt{a^2-b^2}.
\displaystyle AB=2AP=2\sqrt{a^2-b^2}.
\displaystyle \therefore \text{The length of the chord is }2\sqrt{a^2-b^2}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the adjoining figure, }PA\text{ and }PB\text{ are tangents to the circle with centre }O
\displaystyle \text{such that }\angle APB=50^\circ\text{. Write the measure of }\angle OAB\text{.}\hfill\text{[CBSE 2015]} \displaystyle \text{Answer:}
\displaystyle \angle AOB+\angle APB=180^\circ
\displaystyle \therefore \angle AOB=180^\circ-50^\circ=130^\circ.
\displaystyle OA=OB\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OAB=\angle OBA.
\displaystyle \angle OAB=\frac{180^\circ-130^\circ}{2}=25^\circ.
\displaystyle \therefore \angle OAB=25^\circ.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the adjoining figure, }PQ\text{ is a chord of a circle and }PT\text{ is the tangent at }P
\displaystyle \text{such that }\angle QPT=60^\circ\text{. Find }\angle PRQ\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Since }PT\text{ is tangent at }P,\ OP\perp PT.
\displaystyle \therefore \angle OPT=90^\circ.
\displaystyle \angle OPQ=\angle OPT-\angle QPT=90^\circ-60^\circ=30^\circ.
\displaystyle OP=OQ\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OPQ=\angle OQP=30^\circ.
\displaystyle \text{In }\triangle OPQ,
\displaystyle \angle POQ=180^\circ-30^\circ-30^\circ=120^\circ.
\displaystyle \therefore \text{Measure of minor arc }PQ=120^\circ.
\displaystyle \therefore \text{Measure of major arc }PQ=360^\circ-120^\circ=240^\circ.
\displaystyle \text{Since }R\text{ lies on the minor arc }PQ,\ \angle PRQ\text{ subtends the major arc }PQ.
\displaystyle \therefore \angle PRQ=\frac{1}{2}\times240^\circ=120^\circ.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }PQL\text{ and }PRM\text{ are tangents to the circle with centre }O
\displaystyle \text{at the points }Q\text{ and }R\text{ respectively and }S\text{ is a point on the circle. If }\angle SQL=50^\circ
\displaystyle \text{and }\angle SRM=60^\circ,\text{ then find }\angle QSR\text{.}

\displaystyle \text{Answer:}
\displaystyle \text{By the tangent-chord theorem,}
\displaystyle \angle QRS=\angle SQL=50^\circ.
\displaystyle \text{Similarly,}
\displaystyle \angle RQS=\angle SRM=60^\circ.
\displaystyle \text{In }\triangle QRS,
\displaystyle \angle QSR+\angle QRS+\angle RQS=180^\circ
\displaystyle \angle QSR+50^\circ+60^\circ=180^\circ
\displaystyle \therefore \angle QSR=70^\circ.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining figure, }BOA\text{ is a diameter of a circle and the tangent at a point }P
\displaystyle \text{meets }BA\text{ produced at }T\text{. If }\angle PBO=30^\circ,\text{ then find }\angle PTA\text{. Also, show that}
\displaystyle AP=AT\text{.}\hfill\text{[CBSE 2024]}

\displaystyle \text{Answer:}
\displaystyle \text{Since }BA\text{ is a diameter, }\angle BPA=90^\circ\qquad\text{[Angle in a semicircle]}
\displaystyle \text{Also, }B,\ O,\ A\text{ are collinear, so }\angle PBA=\angle PBO=30^\circ.
\displaystyle \text{In }\triangle PBA,
\displaystyle \angle PAB=180^\circ-90^\circ-30^\circ=60^\circ.
\displaystyle \text{Since }BA\text{ is produced to }T,
\displaystyle \angle PAT=180^\circ-\angle PAB=180^\circ-60^\circ=120^\circ.
\displaystyle \text{By the tangent-chord theorem,}
\displaystyle \angle TPA=\angle PBA=30^\circ.
\displaystyle \text{In }\triangle APT,
\displaystyle \angle PTA=180^\circ-120^\circ-30^\circ=30^\circ.
\displaystyle \therefore \angle PTA=\angle TPA=30^\circ.
\displaystyle \therefore AP=AT\qquad\text{[Sides opposite equal angles are equal]}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the adjoining figure, }AB\text{ is diameter of a circle centred at }O\text{. }BC\text{ is tangent to the}
\displaystyle \text{circle at }B\text{. If }OP\text{ bisects the chord }AD\text{ and }\angle AOP=60^\circ,\text{ then find }m\angle C\text{.}
\displaystyle \hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }OP\text{ bisects chord }AD,\ OP\perp AD.
\displaystyle \therefore \angle APO=90^\circ.
\displaystyle \text{In }\triangle AOP,
\displaystyle \angle OAP=180^\circ-90^\circ-60^\circ=30^\circ.
\displaystyle \text{Since }A,\ P,\ D,\ C\text{ are collinear and }A,\ O,\ B\text{ are collinear,}
\displaystyle \angle BAC=\angle OAP=30^\circ.
\displaystyle \text{Since }BC\text{ is tangent at }B,\ OB\perp BC.
\displaystyle \therefore \angle ABC=90^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle C=180^\circ-90^\circ-30^\circ=60^\circ.
\displaystyle \therefore m\angle C=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }XAY\text{ is a tangent to the circle centred at }O\text{. If}
\displaystyle \angle ABO=40^\circ,\text{ then find }m\angle BAY\text{ and }m\angle AOB\text{.}\hfill\text{[CBSE 2022]} \displaystyle \text{Answer:}
\displaystyle OA=OB\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OAB=\angle ABO=40^\circ.
\displaystyle \text{Since }XAY\text{ is tangent at }A,\ OA\perp AY.
\displaystyle \therefore \angle OAY=90^\circ.
\displaystyle \angle OAB+\angle BAY=90^\circ
\displaystyle 40^\circ+\angle BAY=90^\circ
\displaystyle \therefore \angle BAY=50^\circ.
\displaystyle \text{In }\triangle AOB,
\displaystyle \angle AOB=180^\circ-\angle OAB-\angle ABO
\displaystyle =180^\circ-40^\circ-40^\circ=100^\circ.
\displaystyle \therefore m\angle BAY=50^\circ\text{ and }m\angle AOB=100^\circ.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, if tangents }PA\text{ and }PB\text{ drawn from a point }P\text{ to a circle}
\displaystyle \text{with centre }O\text{ are inclined to each other at an angle of }70^\circ,\text{ find }\angle POA\text{.}
\displaystyle \hfill\text{[CBSE 2022]} \displaystyle \text{Answer:}
\displaystyle \angle APB=70^\circ.
\displaystyle \text{The line joining the centre to the external point bisects the angle between the tangents.}
\displaystyle \therefore \angle APO=\frac{70^\circ}{2}=35^\circ.
\displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \therefore \angle OAP=90^\circ.
\displaystyle \text{In }\triangle AOP,
\displaystyle \angle POA=180^\circ-90^\circ-35^\circ=55^\circ.
\displaystyle \therefore \angle POA=55^\circ.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{ }AB\text{ and }AC\text{ are tangents drawn from a point }A\text{ to a circle with centre }O\text{.}
\displaystyle \text{If }\angle BAC=65^\circ,\text{ then find the measure of }\angle BOC\text{.}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB\text{ and }AC\text{ are tangents at }B\text{ and }C\text{ respectively,}
\displaystyle OB\perp AB\qquad\text{and}\qquad OC\perp AC.
\displaystyle \therefore \angle OBA=\angle OCA=90^\circ.
\displaystyle \text{In quadrilateral }ABOC,
\displaystyle \angle BAC+\angle BOC+\angle OBA+\angle OCA=360^\circ
\displaystyle 65^\circ+\angle BOC+90^\circ+90^\circ=360^\circ
\displaystyle \therefore \angle BOC=115^\circ.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the adjoining figure, }PA\text{ is a tangent to the circle drawn from the external point }P
\displaystyle \text{and }PBC\text{ is the secant to the circle with }BC\text{ as diameter. If }\angle AOC=130^\circ,\text{ then find}
\displaystyle \text{the measure of }\angle APB,\text{ where }O\text{ is the centre of the circle.}\hfill\text{[CBSE 2023]} \displaystyle \text{Answer:}
\displaystyle \text{Since }P,\ B,\ O,\ C\text{ are collinear, }\angle AOP+\angle AOC=180^\circ.
\displaystyle \therefore \angle AOP=180^\circ-130^\circ=50^\circ.
\displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \therefore \angle OAP=90^\circ.
\displaystyle \text{In }\triangle AOP,
\displaystyle \angle APB=\angle APO=180^\circ-90^\circ-50^\circ=40^\circ.
\displaystyle \therefore \angle APB=40^\circ.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the adjoining figure, }AB\text{ and }CD\text{ are tangents to a circle centred at }O\text{. Is}
\displaystyle \angle BAC=\angle DCA\text{? Justify your answer.}\hfill\text{[CBSE 2024]} \displaystyle \text{Answer:}
\displaystyle \text{Yes, }\angle BAC=\angle DCA.
\displaystyle OA=OC\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OAC=\angle ACO.\qquad\text{...(i)}
\displaystyle \text{Since }AB\text{ and }CD\text{ are tangents at }A\text{ and }C\text{ respectively,}
\displaystyle OA\perp AB\qquad\text{and}\qquad OC\perp CD.
\displaystyle \therefore \angle BAC=90^\circ-\angle OAC
\displaystyle \text{and }\angle DCA=90^\circ-\angle ACO.
\displaystyle \text{Using (i), }\therefore \angle BAC=\angle DCA.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If two tangents inclined at an angle of }60^\circ\text{ are drawn to a circle of radius}
\displaystyle 3\text{ cm, then find the length of each tangent.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }PA\text{ and }PB\text{ be the tangents from an external point }P\text{ and }O\text{ be the centre.}
\displaystyle \angle APB=60^\circ.
\displaystyle \text{The line joining the centre to the external point bisects the angle between the tangents.}
\displaystyle \therefore \angle APO=\frac{60^\circ}{2}=30^\circ.
\displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \tan30^\circ=\frac{OA}{PA}=\frac{3}{PA}
\displaystyle \frac{1}{\sqrt{3}}=\frac{3}{PA}
\displaystyle \therefore PA=3\sqrt{3}\text{ cm}.
\displaystyle \therefore \text{The length of each tangent is }3\sqrt{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The length of a chord of a bigger circle concentric with a circle of radius }3\text{ cm}
\displaystyle \text{is }8\text{ cm. If the chord is a tangent to the smaller circle, find the radius of the bigger circle.}
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the chord of the bigger circle touching the smaller circle at }P.
\displaystyle \text{Let }O\text{ be the common centre of the two circles.}
\displaystyle OP=3\text{ cm}\qquad\text{and}\qquad AB=8\text{ cm}.
\displaystyle \text{Since }AB\text{ is tangent to the smaller circle at }P,\ OP\perp AB.
\displaystyle \text{The perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore AP=\frac{AB}{2}=\frac{8}{2}=4\text{ cm}.
\displaystyle \text{In right triangle }OAP,
\displaystyle OA^2=OP^2+AP^2
\displaystyle =3^2+4^2=9+16=25
\displaystyle \therefore OA=5\text{ cm}.
\displaystyle \therefore \text{The radius of the bigger circle is }5\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In the adjoining figure, }\triangle PQR\text{ circumscribes the circle. If }PX=10\text{ cm},
\displaystyle QY=8\text{ cm and }RZ=6\text{ cm, then find the perimeter of }\triangle PQR\text{.}\hfill\text{[CBSE 2024]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle PX=PZ=10\text{ cm}
\displaystyle QX=QY=8\text{ cm}
\displaystyle RY=RZ=6\text{ cm}
\displaystyle PQ=PX+XQ=10+8=18\text{ cm}
\displaystyle QR=QY+YR=8+6=14\text{ cm}
\displaystyle PR=PZ+ZR=10+6=16\text{ cm}
\displaystyle \text{Perimeter of }\triangle PQR=PQ+QR+PR
\displaystyle =18+14+16=48\text{ cm}.
\displaystyle \therefore \text{The perimeter of }\triangle PQR\text{ is }48\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A person is standing at }P\text{ outside a circular ground at a distance of }26\text{ m}
\displaystyle \text{from the centre of the ground. He found that his distances from the points }A\text{ and }B\text{ on the}
\displaystyle \text{ground are }10\text{ m }(PA\text{ and }PB\text{ are tangents to the circle). Find the radius of the circular}
\displaystyle \text{ground.}\hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle OP=26\text{ m}\qquad\text{and}\qquad PA=10\text{ m}.
\displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \text{In right triangle }OAP,
\displaystyle OP^2=OA^2+AP^2
\displaystyle OA^2=OP^2-AP^2
\displaystyle =26^2-10^2=676-100=576
\displaystyle \therefore OA=\sqrt{576}=24\text{ m}.
\displaystyle \therefore \text{The radius of the circular ground is }24\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{At point }A\text{ on the diameter }AB\text{ of a circle of radius }10\text{ cm, tangent }XAY
\displaystyle \text{is drawn to the circle. Find the length of the chord }CD\text{ parallel to }XY\text{ at a distance of }16\text{ cm}
\displaystyle \text{from }A\text{.}\hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Let }CD\text{ intersect the diameter }AB\text{ at }P.
\displaystyle \text{Since }XY\text{ is tangent at }A,\ OA\perp XY.
\displaystyle \text{Also, }CD\parallel XY.
\displaystyle \therefore OP\perp CD.
\displaystyle \text{The perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore CP=PD.
\displaystyle AP=16\text{ cm}\qquad\text{and}\qquad AO=OC=10\text{ cm}.
\displaystyle OP=AP-AO=16-10=6\text{ cm}.
\displaystyle \text{In right triangle }OPC,
\displaystyle OC^2=OP^2+PC^2
\displaystyle PC^2=10^2-6^2=100-36=64
\displaystyle \therefore PC=8\text{ cm}.
\displaystyle CD=2PC=2\times8=16\text{ cm}.
\displaystyle \therefore \text{The length of the chord }CD\text{ is }16\text{ cm}.
\displaystyle \\

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