\displaystyle \textbf{Question 1: }\text{If }PT\text{ is a tangent at }T\text{ to a circle whose centre is }O\text{ and}
\displaystyle OP=17\text{ cm},\ OT=8\text{ cm, find the length of the tangent segment }PT\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the radius through the point of contact is perpendicular to the tangent,}
\displaystyle OT\perp PT.
\displaystyle \therefore \triangle OTP\text{ is right-angled at }T.
\displaystyle OP^2=OT^2+PT^2
\displaystyle PT^2=OP^2-OT^2
\displaystyle =17^2-8^2=289-64=225
\displaystyle \therefore PT=\sqrt{225}=15\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A point }P\text{ is }26\text{ cm away from the centre }O\text{ of a circle and the}
\displaystyle \text{length }PT\text{ of the tangent drawn from }P\text{ to the circle is }10\text{ cm. Find its radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle }OT=r\text{ cm.}
\displaystyle \text{Since the radius through the point of contact is perpendicular to the tangent,}
\displaystyle OT\perp PT.
\displaystyle \therefore \triangle OTP\text{ is right-angled at }T.
\displaystyle OP^2=OT^2+PT^2
\displaystyle r^2=OP^2-PT^2
\displaystyle =26^2-10^2=676-100=576
\displaystyle \therefore r=\sqrt{576}=24\text{ cm}.
\displaystyle \therefore \text{The radius of the circle is }24\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }PA\text{ and }PB\text{ are tangents to the circle from an}
\displaystyle \text{external point }P\text{. }CD\text{ is another tangent touching the circle at }Q.
\displaystyle \text{If }PA=12\text{ cm,} \ QC=QD=3\text{ cm, then find }PC+PD\text{.}\hfill\text{[CBSE 2017]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore PA=PB=12\text{ cm}.
\displaystyle CA=CQ=3\text{ cm}\qquad\text{and}\qquad DB=DQ=3\text{ cm}
\displaystyle PA=PC+CA
\displaystyle \therefore PC=PA-CA=12-3=9\text{ cm}.
\displaystyle PB=PD+DB
\displaystyle \therefore PD=PB-DB=12-3=9\text{ cm}.
\displaystyle \therefore PC+PD=9+9=18\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }AB,\ AC,\ PQ\text{ are tangents in the adjoining figure and }
\displaystyle AB=5\text{ cm, find the} \ \text{perimeter of }\triangle APQ\text{.}\hfill\text{[CBSE 2000]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore AB=AC=5\text{ cm}.
\displaystyle PB=PX\qquad\text{and}\qquad QC=QX
\displaystyle AP=AB-PB\qquad\text{and}\qquad AQ=AC-QC
\displaystyle PQ=PX+XQ=PB+QC
\displaystyle \text{Perimeter of }\triangle APQ=AP+AQ+PQ
\displaystyle =(AB-PB)+(AC-QC)+(PB+QC)
\displaystyle =AB+AC=5+5=10\text{ cm}.
\displaystyle \therefore \text{The perimeter of }\triangle APQ\text{ is }10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the adjoining figure, }PQ\text{ is tangent at a point }R\text{ of the circle with centre}
\displaystyle O\text{.} \ \text{If }\angle TRQ=30^\circ,\text{ find }\angle PRS\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Since }ST\text{ passes through the centre }O,\ ST\text{ is a diameter.}
\displaystyle \therefore \angle SRT=90^\circ\qquad\text{[Angle in a semicircle]}
\displaystyle \text{By the tangent-chord theorem,}
\displaystyle \angle RST=\angle TRQ=30^\circ.
\displaystyle \text{In }\triangle SRT,
\displaystyle \angle SRT+\angle RST+\angle RTS=180^\circ
\displaystyle 90^\circ+30^\circ+\angle RTS=180^\circ
\displaystyle \therefore \angle RTS=60^\circ.
\displaystyle \text{Again, by the tangent-chord theorem,}
\displaystyle \angle PRS=\angle RTS=60^\circ.
\displaystyle \therefore \angle PRS=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, a circle touches all the four sides of a quadrilateral }
\displaystyle ABCD\text{ with} \ AB=6\text{ cm},\ BC=7\text{ cm and }CD=4\text{ cm. Find }AD\text{.}\hfill\text{[CBSE 2002]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle AP=AS,\quad BP=BQ,\quad CQ=CR,\quad DR=DS
\displaystyle AB+CD=(AP+PB)+(CR+RD)
\displaystyle =(AS+BQ)+(CQ+DS)
\displaystyle =(AS+DS)+(BQ+CQ)
\displaystyle =AD+BC
\displaystyle \therefore AB+CD=AD+BC
\displaystyle 6+4=AD+7
\displaystyle \therefore AD=10-7=3\text{ cm}.
\displaystyle \therefore \text{The length of }AD\text{ is }3\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the adjoining figure, there are two concentric circles with centre }
\displaystyle O\text{ of radii }5\text{ cm} \ \text{and }3\text{ cm. From an external point }P,\text{ tangents }PA\text{ and }PB\text{ are}
\displaystyle \text{drawn to these circles. If }AP=12\text{ cm, find the length of }BP\text{.}\hfill\text{[CBSE 2010, 2012, 2016]} \displaystyle \text{Answer:}
\displaystyle \text{Since the radius through the point of contact is perpendicular to the tangent,}
\displaystyle OA\perp PA.
\displaystyle \text{In right triangle }OAP,
\displaystyle OP^2=OA^2+AP^2
\displaystyle =5^2+12^2=25+144=169
\displaystyle \therefore OP=13\text{ cm}.
\displaystyle \text{Similarly, }OB\perp PB\text{ and }OB=3\text{ cm}.
\displaystyle \text{In right triangle }OBP,
\displaystyle OP^2=OB^2+BP^2
\displaystyle BP^2=OP^2-OB^2
\displaystyle =13^2-3^2=169-9=160
\displaystyle \therefore BP=\sqrt{160}=4\sqrt{10}\text{ cm}.
\displaystyle \therefore \text{The length of }BP\text{ is }4\sqrt{10}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(i) In the adjoining figure (i), a circle is inscribed in a quadrilateral }ABCD\text{ in which }
\displaystyle \angle B=90^\circ\text{. If} \ AD=23\text{ cm},\ AB=29\text{ cm and }
\displaystyle DS=5\text{ cm, find the radius of the inscribed circle.} \ \hfill\text{[CBSE 2023]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle DR=DS=5\text{ cm}.
\displaystyle AR=AD-DR=23-5=18\text{ cm}.
\displaystyle AQ=AR=18\text{ cm}\qquad\text{[Tangents from }A\text{]}
\displaystyle BQ=AB-AQ=29-18=11\text{ cm}.
\displaystyle \text{Since }\angle B=90^\circ\text{ and the circle touches }AB\text{ and }BC,
\displaystyle BQ=BP=\text{radius of the circle}.
\displaystyle \therefore \text{The radius of the inscribed circle is }11\text{ cm}.
\displaystyle \\

\displaystyle \text{(ii) In the adjoining figure (ii), a circle with centre }O\text{ and radius }8\text{ cm is inscribed in a }
\displaystyle \text{quadrilateral } ABCD\text{ in which }P,Q,R,S\text{ are the points of contact. If }
\displaystyle AD\perp DC,\   BC=30\text{ cm} \ \text{and }BS=24\text{ cm, then find the length of }DC\text{.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle BR=BS=24\text{ cm}.
\displaystyle CR=BC-BR=30-24=6\text{ cm}.
\displaystyle CQ=CR=6\text{ cm}\qquad\text{[Tangents from }C\text{]}
\displaystyle \text{Since }AD\perp DC\text{ and the radius of the circle is }8\text{ cm,}
\displaystyle DP=DQ=8\text{ cm}.
\displaystyle DC=DQ+QC=8+6=14\text{ cm}.
\displaystyle \therefore \text{The length of }DC\text{ is }14\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }AB\text{ is a chord of length }16\text{ cm of a circle of radius }
\displaystyle 10\text{ cm.} \ \text{The tangents at }A\text{ and }B\text{ intersect at a point }P\text{. Find the length of }PA\text{.}
\displaystyle \hfill\text{[CBSE 2010]} \displaystyle \text{Answer:}
\displaystyle \text{Let }OM\perp AB\text{, where }M\text{ is the midpoint of chord }AB.
\displaystyle AM=BM=\frac{AB}{2}=\frac{16}{2}=8\text{ cm}.
\displaystyle \text{In right triangle }OMA,
\displaystyle OA^2=OM^2+AM^2
\displaystyle 10^2=OM^2+8^2
\displaystyle OM^2=100-64=36
\displaystyle \therefore OM=6\text{ cm}.
\displaystyle \text{Let }PM=x\text{ cm. Therefore, }OP=x+6\text{ cm}.
\displaystyle \text{In right triangle }PMA,
\displaystyle PA^2=PM^2+AM^2=x^2+64\qquad\text{...(i)}
\displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \text{In right triangle }OAP,
\displaystyle OP^2=OA^2+PA^2
\displaystyle (x+6)^2=100+x^2+64\qquad\text{[Using (i)]}
\displaystyle x^2+12x+36=x^2+164
\displaystyle 12x=128\Rightarrow x=\frac{32}{3}.
\displaystyle PA^2=x^2+64=\left(\frac{32}{3}\right)^2+64
\displaystyle =\frac{1024}{9}+\frac{576}{9}=\frac{1600}{9}
\displaystyle \therefore PA=\frac{40}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two concentric circles are of diameters }30\text{ cm and }18\text{ cm. Find the}
\displaystyle \text{length of the chord of the larger circle which touches the smaller circle.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the larger circle}=\frac{30}{2}=15\text{ cm}.
\displaystyle \text{Radius of the smaller circle}=\frac{18}{2}=9\text{ cm}.
\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{The perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore AP=PB.
\displaystyle \text{In right triangle }OAP,
\displaystyle OA^2=OP^2+AP^2
\displaystyle 15^2=9^2+AP^2
\displaystyle AP^2=225-81=144
\displaystyle \therefore AP=12\text{ cm}.
\displaystyle \therefore AB=2AP=2\times12=24\text{ cm}.
\displaystyle \therefore \text{The length of the chord is }24\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If from any point on the common chord of two intersecting circles, tangents}
\displaystyle \text{be drawn to the circles, prove that they are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two circles intersect at }A\text{ and }B,\text{ and let }P\text{ be a point on }AB\text{ produced.}
\displaystyle \text{Let }PT\text{ and }PS\text{ be tangents from }P\text{ to the two circles respectively.}
\displaystyle \text{For the first circle, by the tangent-secant theorem,}
\displaystyle PT^2=PA\cdot PB.\qquad\text{...(i)}
\displaystyle \text{For the second circle, by the tangent-secant theorem,}
\displaystyle PS^2=PA\cdot PB.\qquad\text{...(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle PT^2=PS^2
\displaystyle \therefore PT=PS.
\displaystyle \therefore \text{The tangents drawn from }P\text{ to the two circles are equal.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Two circles touch externally at a point }P\text{. From a point }
\displaystyle T\text{ on the tangent at }P, \ \text{tangents }TQ\text{ and }TR\text{ are drawn to the circles with points}
\displaystyle \text{of contact } Q\text{ and }R \ \text{respectively. Prove that }TQ=TR\text{.} \displaystyle \text{Answer:}
\displaystyle \text{For the left circle, }TQ\text{ and }TP\text{ are tangents drawn from the external point }T.
\displaystyle \therefore TQ=TP.\qquad\text{...(i)}
\displaystyle \text{For the right circle, }TR\text{ and }TP\text{ are tangents drawn from the external point }T.
\displaystyle \therefore TR=TP.\qquad\text{...(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle TQ=TR.
\displaystyle \therefore \text{The tangents }TQ\text{ and }TR\text{ are equal.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }PA\text{ and }PB\text{ are tangents from an external point }P
\displaystyle \text{ to a circle} \ \text{with centre }O\text{. }LN\text{ touches the circle at }M.
\displaystyle \text{Prove that } PL+LM=PN+MN\text{.}\ \hfill\text{[CBSE 2010]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore LA=LM\qquad\text{[Tangents from }L\text{]}
\displaystyle NB=NM\qquad\text{[Tangents from }N\text{]}
\displaystyle PA=PB\qquad\text{[Tangents from }P\text{]}
\displaystyle PA=PL+LA=PL+LM
\displaystyle PB=PN+NB=PN+MN
\displaystyle \text{Since }PA=PB,
\displaystyle \therefore PL+LM=PN+MN.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining figure, }ABC\text{ is a right triangle right-angled at }B
\displaystyle \text{ such that} \ BC=6\text{ cm and }AB=8\text{ cm. Find the radius of its incircle.}\hfill\text{[CBSE 2002]} \displaystyle \text{Answer:}
\displaystyle \text{In right triangle }ABC,
\displaystyle AC^2=AB^2+BC^2
\displaystyle =8^2+6^2=64+36=100
\displaystyle \therefore AC=10\text{ cm}.
\displaystyle \text{For a right triangle, the radius }r\text{ of the incircle is}
\displaystyle r=\frac{AB+BC-AC}{2}
\displaystyle =\frac{8+6-10}{2}=\frac{4}{2}=2\text{ cm}.
\displaystyle \therefore \text{The radius of the incircle is }2\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the adjoining figure, }BDC\text{ is a tangent to the given circle at point }
\displaystyle D\text{ such that} \ BD=30\text{ cm and }CD=7\text{ cm. The other tangents }BE\text{ and }CF
\displaystyle \text{ are drawn respectively}  \ \text{from }B\text{ and }C\text{ to the circle and meet when produced at }A,
\displaystyle \text{ making }\triangle BAC \ \text{a right triangle. Calculate (i) }AF\text{ (ii) radius of the circle.} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore BE=BD=30\text{ cm}\qquad\text{and}\qquad CF=CD=7\text{ cm}.
\displaystyle \text{Let the radius of the circle be }r\text{ cm}.
\displaystyle \text{Since }\angle BAC=90^\circ,\ AE=AF=r\text{ cm}.
\displaystyle AB=AE+EB=r+30
\displaystyle AC=AF+FC=r+7
\displaystyle BC=BD+DC=30+7=37\text{ cm}.
\displaystyle \text{Applying Pythagoras theorem in right triangle }ABC,
\displaystyle BC^2=AB^2+AC^2
\displaystyle 37^2=(r+30)^2+(r+7)^2
\displaystyle 1369=r^2+60r+900+r^2+14r+49
\displaystyle 2r^2+74r-420=0
\displaystyle r^2+37r-210=0
\displaystyle (r+42)(r-5)=0
\displaystyle \therefore r=5\text{ cm}\qquad\text{[Rejecting }r=-42\text{]}
\displaystyle \text{(i) }AF=r=5\text{ cm}.
\displaystyle \text{(ii) Radius of the circle}=5\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, the tangent at a point }C\text{ of a circle and a diameter }
\displaystyle AB\text{ when} \ \text{extended intersect at }P\text{. If }\angle PCA=110^\circ,\text{ find }\angle CBA\text{ and }\angle BCO\text{.}
\displaystyle \hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Since }AB\text{ is a diameter of the circle,}
\displaystyle \angle ACB=90^\circ\qquad\text{[Angle in a semicircle]}
\displaystyle \angle PCA=\angle PCB+\angle BCA
\displaystyle 110^\circ=\angle PCB+90^\circ
\displaystyle \therefore \angle PCB=20^\circ.
\displaystyle \text{Since }PC\text{ is tangent to the circle at }C\text{ and }OC\text{ is a radius,}
\displaystyle OC\perp PC.
\displaystyle \therefore \angle OCP=90^\circ.
\displaystyle \angle OCB+\angle BCP=90^\circ
\displaystyle \angle OCB+20^\circ=90^\circ
\displaystyle \therefore \angle BCO=\angle OCB=70^\circ.
\displaystyle \text{In }\triangle OBC,\ OB=OC\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OBC=\angle BCO=70^\circ.
\displaystyle \text{Since }A,\ O,\ B\text{ are collinear, }\angle CBA=\angle CBO=70^\circ.
\displaystyle \therefore \angle CBA=70^\circ\text{ and }\angle BCO=70^\circ.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }PQ\text{ is a tangent from an external point }P
\displaystyle \text{ to a circle with} \ \text{centre }O\text{ and }OP\text{ cuts the circle at }T\text{ and }QOR\text{ is a diameter. If }
\displaystyle \angle POR=130^\circ \ \text{and }S\text{ is a point on the circle, find }\angle 1+\angle 2\text{.}\hfill\text{[CBSE 2017]} \displaystyle \text{Answer:}
\displaystyle \text{Since }QOR\text{ is a diameter, }OQ\text{ and }OR\text{ are opposite rays.}
\displaystyle \therefore \angle QOP=180^\circ-\angle POR=180^\circ-130^\circ=50^\circ.
\displaystyle \text{Since }PQ\text{ is tangent at }Q,\ OQ\perp PQ.
\displaystyle \therefore \angle OQP=90^\circ.
\displaystyle \text{In }\triangle OPQ,
\displaystyle \angle 1+\angle QOP+\angle OQP=180^\circ
\displaystyle \angle 1+50^\circ+90^\circ=180^\circ
\displaystyle \therefore \angle 1=40^\circ.
\displaystyle \text{Since }P,\ T,\ O\text{ are collinear, }\angle ROT=\angle ROP=130^\circ.
\displaystyle \angle 2=\angle RST=\frac{1}{2}\angle ROT
\displaystyle =\frac{1}{2}\times130^\circ=65^\circ.
\displaystyle \therefore \angle 1+\angle 2=40^\circ+65^\circ=105^\circ.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the adjoining figure, a }\triangle ABC\text{ is drawn to circumscribe}
\displaystyle \text{a circle of radius }4\text{ cm} \ \text{such that the segments }BD\text{ and }DC\text{ are of lengths }8\text{ cm and }
\displaystyle 6\text{ cm respectively. Find} \ \text{the lengths of sides }AB\text{ and }AC,\text{ when area of }
\displaystyle \triangle ABC\text{ is }84\text{ cm}^2\text{.}  \ \hfill\text{[CBSE 2015, CBSE 2023]} \displaystyle \text{Answer:}
\displaystyle \text{Let }s\text{ be the semi-perimeter of }\triangle ABC.
\displaystyle \text{Area of }\triangle ABC=r\times s
\displaystyle 84=4s
\displaystyle \therefore s=21\text{ cm}.
\displaystyle \therefore \text{Perimeter of }\triangle ABC=2s=42\text{ cm}.
\displaystyle BC=BD+DC=8+6=14\text{ cm}.
\displaystyle \therefore AB+AC=42-14=28\text{ cm}.\qquad\text{...(i)}
\displaystyle \text{Let the circle touch }AB\text{ and }AC\text{ at }P\text{ and }Q\text{ respectively.}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle BP=BD=8\text{ cm},\qquad CQ=CD=6\text{ cm}
\displaystyle \text{and let }AP=AQ=x\text{ cm}.
\displaystyle AB=x+8\qquad\text{and}\qquad AC=x+6
\displaystyle \text{Using (i),}
\displaystyle (x+8)+(x+6)=28
\displaystyle 2x+14=28
\displaystyle \therefore x=7\text{ cm}.
\displaystyle AB=7+8=15\text{ cm}
\displaystyle AC=7+6=13\text{ cm}
\displaystyle \therefore AB=15\text{ cm and }AC=13\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the adjoining figure, }AB\text{ is a diameter of a circle with centre }O\text{ and }AT
\displaystyle \text{is a tangent. If }\angle AOQ=58^\circ,\text{ find }\angle ATQ\text{.}\hfill\text{[CBSE 2015]} \displaystyle \text{Answer:}
\displaystyle \angle AOQ=58^\circ
\displaystyle \text{The angle subtended by an arc at the centre is twice the angle subtended by it}
\displaystyle \text{at any point on the remaining part of the circle.}
\displaystyle \therefore \angle AOQ=2\angle ABQ
\displaystyle 58^\circ=2\angle ABQ
\displaystyle \therefore \angle ABQ=29^\circ.
\displaystyle \text{Since }AT\text{ is tangent at }A\text{ and }OA\text{ is a radius, }OA\perp AT.
\displaystyle \text{Also, }A,\ O,\ B\text{ are collinear.}
\displaystyle \therefore \angle BAT=90^\circ.
\displaystyle \text{Since }B,\ Q,\ T\text{ are collinear, }\angle ABT=\angle ABQ=29^\circ.
\displaystyle \text{In }\triangle ABT,
\displaystyle \angle BAT+\angle ABT+\angle ATB=180^\circ
\displaystyle 90^\circ+29^\circ+\angle ATB=180^\circ
\displaystyle \therefore \angle ATB=61^\circ.
\displaystyle \text{Since }B,\ Q,\ T\text{ are collinear, }\angle ATQ=\angle ATB.
\displaystyle \therefore \angle ATQ=61^\circ.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the adjoining figure, }OQ:PQ=3:4\text{ and perimeter of }
\displaystyle \triangle POQ=60\text{ cm.} \ \text{Determine }PQ,\ QR\text{ and }OP\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Since }PQ\text{ is tangent at }Q\text{ and }OQ\text{ is a radius, }OQ\perp PQ.
\displaystyle \therefore \triangle POQ\text{ is right-angled at }Q.
\displaystyle \text{Given, }OQ:PQ=3:4.
\displaystyle \text{Let }OQ=3x\text{ cm and }PQ=4x\text{ cm}.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle OP=\sqrt{(3x)^2+(4x)^2}=5x\text{ cm}.
\displaystyle \text{Perimeter of }\triangle POQ=60\text{ cm}
\displaystyle 3x+4x+5x=60
\displaystyle 12x=60
\displaystyle \therefore x=5.
\displaystyle PQ=4x=4\times5=20\text{ cm}.
\displaystyle OQ=3x=3\times5=15\text{ cm}.
\displaystyle OP=5x=5\times5=25\text{ cm}.
\displaystyle \text{Since }QR\text{ is a diameter, }QR=2OQ=2\times15=30\text{ cm}.
\displaystyle \therefore PQ=20\text{ cm},\ QR=30\text{ cm and }OP=25\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In the adjoining figure, there are two concentric circles with centre }O\text{. }
\displaystyle PRT\text{ and} \ PQS\text{ are tangents to the inner circle from a point }P\text{ lying on the outer circle. If }
\displaystyle PR=5\text{ cm,}  \ \text{find the length of }PS\text{.}\hfill\text{[CBSE 2017]} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore PQ=PR=5\text{ cm}.
\displaystyle \text{Since }PQS\text{ is tangent to the inner circle at }Q,\ OQ\perp PS.
\displaystyle \text{Also, }PS\text{ is a chord of the outer circle.}
\displaystyle \text{The perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore PQ=QS=5\text{ cm}.
\displaystyle PS=PQ+QS=5+5=10\text{ cm}.
\displaystyle \therefore \text{The length of }PS\text{ is }10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }PA\text{ and }PB\text{ are tangents from an outside point }P\text{ such that }
\displaystyle PA=10\text{ cm} \ \text{and }\angle APB=60^\circ,\text{ find the length of chord }AB\text{.}\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore PA=PB=10\text{ cm}.
\displaystyle \therefore \angle PAB=\angle PBA\qquad\text{[Angles opposite equal sides]}
\displaystyle \text{In }\triangle PAB,
\displaystyle \angle PAB+\angle PBA+\angle APB=180^\circ
\displaystyle 2\angle PAB+60^\circ=180^\circ
\displaystyle \therefore \angle PAB=\angle PBA=60^\circ.
\displaystyle \therefore \triangle PAB\text{ is an equilateral triangle.}
\displaystyle \therefore AB=PA=10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{From an external point }P,\text{ tangents }PA=PB\text{ are drawn to a circle}
\displaystyle \text{with centre }O\text{.} \ \text{If }\angle PAB=50^\circ,\text{ then find }\angle AOB\text{.}\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:} \displaystyle PA=PB\qquad\text{[Tangents from the same external point]}
\displaystyle \therefore \angle PAB=\angle PBA=50^\circ.
\displaystyle \text{In }\triangle PAB,
\displaystyle \angle APB=180^\circ-50^\circ-50^\circ=80^\circ.
\displaystyle \text{Since }OA\perp PA\text{ and }OB\perp PB,
\displaystyle \angle OAP=\angle OBP=90^\circ.
\displaystyle \text{In quadrilateral }OAPB,
\displaystyle \angle AOB+\angle APB+90^\circ+90^\circ=360^\circ
\displaystyle \angle AOB+80^\circ=180^\circ
\displaystyle \therefore \angle AOB=100^\circ.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The lengths of three consecutive sides of a quadrilateral circumscribing}
\displaystyle \text{a circle are }4\text{ cm},\ 5\text{ cm and }7\text{ cm respectively. Determine the length of the fourth side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the fourth side be }x\text{ cm}.
\displaystyle \text{In a quadrilateral circumscribing a circle, the sums of opposite sides are equal.}
\displaystyle \therefore 4+7=5+x
\displaystyle 11=5+x
\displaystyle \therefore x=6\text{ cm}.
\displaystyle \therefore \text{The length of the fourth side is }6\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In the adjoining figure, }O\text{ is the centre of the circle and }
\displaystyle BCD\text{ is tangent to it at }C\text{.}  \text{Prove that }\angle BAC+\angle ACD=90^\circ\text{.}\hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Since }B,\ P,\ O,\ A\text{ are collinear and }O\text{ is the centre, }AP\text{ is a diameter.}
\displaystyle \therefore \angle PCA=90^\circ\qquad\text{[Angle in a semicircle]}
\displaystyle \text{In }\triangle APC,
\displaystyle \angle PAC+\angle APC=90^\circ.\qquad\text{...(i)}
\displaystyle \text{Since }B,\ P,\ A\text{ are collinear,}
\displaystyle \angle BAC=\angle PAC.\qquad\text{...(ii)}
\displaystyle \text{By the tangent-chord theorem,}
\displaystyle \angle ACD=\angle APC.\qquad\text{...(iii)}
\displaystyle \text{Using (ii) and (iii) in (i),}
\displaystyle \therefore \angle BAC+\angle ACD=90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In the adjoining figure, }PB\text{ is a tangent to the circle with centre }
\displaystyle O\text{ at }B\text{. }AB \ \text{is a chord of the circle of length }24\text{ cm and at a distance of }5\text{ cm from}
\displaystyle \text{the centre. If the length }PB\text{ of the tangent is }20\text{ cm, find the length of }OP\text{.}\hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Since }OM\perp AB,\text{ the perpendicular from the centre to a chord bisects the chord.}
\displaystyle \therefore MB=\frac{AB}{2}=\frac{24}{2}=12\text{ cm}.
\displaystyle \text{In right triangle }OMB,
\displaystyle OB^2=OM^2+MB^2
\displaystyle =5^2+12^2=25+144=169
\displaystyle \therefore OB=13\text{ cm}.
\displaystyle \text{Since }PB\text{ is tangent at }B,\ OB\perp PB.
\displaystyle \text{In right triangle }OBP,
\displaystyle OP^2=OB^2+PB^2
\displaystyle =13^2+20^2=169+400=569
\displaystyle \therefore OP=\sqrt{569}\text{ cm}.
\displaystyle \therefore \text{The length of }OP\text{ is }\sqrt{569}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Rectangle }ABCD\text{ circumscribes a circle of radius }10\text{ cm. Prove that }
\displaystyle ABCD \ \text{is a square. Hence, find the perimeter of }ABCD\text{.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{For a quadrilateral circumscribing a circle, the sums of opposite sides are equal.}
\displaystyle \therefore AB+CD=BC+AD.
\displaystyle \text{Since }ABCD\text{ is a rectangle,}
\displaystyle AB=CD\qquad\text{and}\qquad BC=AD.
\displaystyle \therefore AB+AB=BC+BC
\displaystyle 2AB=2BC
\displaystyle \therefore AB=BC.
\displaystyle \text{Thus, adjacent sides of the rectangle are equal.}
\displaystyle \therefore ABCD\text{ is a square.}
\displaystyle \text{The distance between two opposite sides of the square equals the diameter of the circle.}
\displaystyle \therefore \text{Side of the square}=2r=2\times10=20\text{ cm}.
\displaystyle \text{Perimeter of }ABCD=4\times20=80\text{ cm}.
\displaystyle \therefore \text{The perimeter of }ABCD\text{ is }80\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{In the adjoining figure, }TP\text{ and }TQ\text{ are tangents drawn to a}
\displaystyle \text{circle with centre }O\text{.} \ \text{If }\angle OPQ=15^\circ\text{ and }\angle PTQ=\theta, \text{ then find the value of } \\ \sin 2\theta\text{.} \hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle OP=OQ\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \angle OPQ=\angle OQP=15^\circ.
\displaystyle \text{In }\triangle OPQ,
\displaystyle \angle POQ=180^\circ-15^\circ-15^\circ=150^\circ.
\displaystyle \text{The angle between two tangents from an external point and the angle subtended}
\displaystyle \text{by the points of contact at the centre are supplementary.}
\displaystyle \therefore \angle PTQ+\angle POQ=180^\circ
\displaystyle \theta+150^\circ=180^\circ
\displaystyle \therefore \theta=30^\circ.
\displaystyle \therefore \sin2\theta=\sin(2\times30^\circ)=\sin60^\circ=\frac{\sqrt{3}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In the adjoining figure, equal circles with centres }O\text{ and }O'
\displaystyle \text{ touch each other at }X\text{.}  \ OO'\text{ produced meets the circle with centre }O'\text{ at }A\text{. }AC\text{ is a tangent}
\displaystyle \text{to the circle whose centre is }O\text{. }O'D\text{ is perpendicular to }AC\text{. Find the value of }\frac{DO'}{CO}\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Let the radius of each circle be }r.
\displaystyle AO'=O'X=XO=OC=r.
\displaystyle AO=AO'+O'X+XO=r+r+r=3r.
\displaystyle \text{Since }O'D\perp AC,\ \angle ADO'=90^\circ.
\displaystyle \text{Since }AC\text{ is tangent at }C,\ OC\perp AC.
\displaystyle \therefore \angle ACO=90^\circ.
\displaystyle \text{Also, }\angle DAO'=\angle CAO.
\displaystyle \therefore \triangle ADO'\sim\triangle ACO\qquad\text{[AA similarity]}
\displaystyle \therefore \frac{DO'}{CO}=\frac{AO'}{AO}
\displaystyle =\frac{r}{3r}=\frac{1}{3}.
\displaystyle \therefore \frac{DO'}{CO}=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{In the adjoining figure, }BC\text{ is a tangent to the circle with centre }O\text{. }
\displaystyle OE\text{ bisects }AP\text{.} \ \text{Prove that }\triangle AEO\sim\triangle ABC\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Since }OE\text{ bisects the chord }AP,\ E\text{ is the midpoint of }AP.
\displaystyle \text{The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.}
\displaystyle \therefore OE\perp AP
\displaystyle \therefore \angle AEO=90^\circ.\qquad\text{...(i)}
\displaystyle \text{Since }BC\text{ is tangent to the circle at }B,\ OB\perp BC.
\displaystyle \text{Also, }A,\ O,\ B\text{ are collinear.}
\displaystyle \therefore \angle ABC=90^\circ.\qquad\text{...(ii)}
\displaystyle \text{Since }A,\ E,\ C\text{ are collinear and }A,\ O,\ B\text{ are collinear,}
\displaystyle \angle EAO=\angle CAB.\qquad\text{...(iii)}
\displaystyle \text{From (i), (ii) and (iii),}
\displaystyle \angle AEO=\angle ABC\quad\text{and}\quad\angle EAO=\angle CAB.
\displaystyle \therefore \triangle AEO\sim\triangle ABC\qquad\text{[AA similarity criterion]}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In the adjoining figure, }PO\perp QO\text{. The tangents to the circle at }P\text{ and }Q
\displaystyle \text{intersect at a point }T\text{. Prove that }PQ\text{ and }OT\text{ are right bisectors of each other.} \displaystyle \text{Answer:}
\displaystyle \text{Since }PT\text{ is tangent to the circle at }P,\ OP\perp PT.
\displaystyle \text{Also, }QT\text{ is tangent to the circle at }Q,\ OQ\perp QT.
\displaystyle \text{Given, }OP\perp OQ.
\displaystyle \therefore PT\parallel OQ\qquad\text{and}\qquad QT\parallel OP.
\displaystyle \therefore OPTQ\text{ is a parallelogram.}
\displaystyle \text{Also, }\angle POQ=90^\circ.
\displaystyle \therefore OPTQ\text{ is a rectangle.}
\displaystyle OP=OQ\qquad\text{[Radii of the same circle]}
\displaystyle \therefore \text{the adjacent sides of rectangle }OPTQ\text{ are equal.}
\displaystyle \therefore OPTQ\text{ is a square.}
\displaystyle \text{The diagonals of a square bisect each other at right angles.}
\displaystyle \therefore PQ\text{ and }OT\text{ bisect each other at right angles.}
\displaystyle \therefore PQ\text{ and }OT\text{ are right bisectors of each other.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{In the adjoining figure, }O\text{ is the centre of the circle and }
\displaystyle BCD\text{ is tangent to it at }C\text{.} \ \text{Prove that }\angle BAC+\angle ACD=90^\circ\text{.}\hfill\text{[CBSE 2023]} \displaystyle \text{Answer:}
\displaystyle \text{Since }P,\ O,\ A\text{ are collinear and }O\text{ is the centre, }PA\text{ is a diameter.}
\displaystyle \therefore \angle PCA=90^\circ\qquad\text{[Angle in a semicircle]}
\displaystyle \text{In }\triangle APC,
\displaystyle \angle PAC+\angle APC=90^\circ.\qquad\text{...(i)}
\displaystyle \text{Since }B,\ P,\ A\text{ are collinear,}
\displaystyle \angle BAC=\angle PAC.\qquad\text{...(ii)}
\displaystyle \text{By the tangent-chord theorem,}
\displaystyle \angle ACD=\angle APC.\qquad\text{...(iii)}
\displaystyle \text{Using (ii) and (iii) in (i),}
\displaystyle \therefore \angle BAC+\angle ACD=90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{From a point }P\text{ two tangents }PA\text{ and }PB\text{ are drawn to a}
\displaystyle \text{circle with centre }O\text{.} \ \text{If }OP=2r,\text{ show that }\triangle PAB\text{ is equilateral.}\hfill\text{[CBSE 2008, 11, 12]}
\displaystyle \text{Answer:} \displaystyle \text{Since }PA\text{ is tangent at }A,\ OA\perp PA.
\displaystyle \text{In right triangle }OAP,\ OA=r\text{ and }OP=2r.
\displaystyle \sin\angle OPA=\frac{OA}{OP}=\frac{r}{2r}=\frac{1}{2}
\displaystyle \therefore \angle OPA=30^\circ.
\displaystyle \text{The line joining the centre to the external point bisects the angle between the tangents.}
\displaystyle \therefore \angle OPA=\angle OPB=30^\circ.
\displaystyle \therefore \angle APB=30^\circ+30^\circ=60^\circ.
\displaystyle PA=PB\qquad\text{[Tangents from the same external point]}
\displaystyle \therefore \angle PAB=\angle PBA.
\displaystyle \text{In }\triangle PAB,
\displaystyle \angle PAB+\angle PBA+\angle APB=180^\circ
\displaystyle 2\angle PAB+60^\circ=180^\circ
\displaystyle \therefore \angle PAB=\angle PBA=60^\circ.
\displaystyle \therefore \triangle PAB\text{ is equilateral.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{A triangle }PQR\text{ is drawn to circumscribe a circle of radius }8\text{ cm such that}
\displaystyle \text{the segments }QT\text{ and }TR,\text{ into which }QR\text{ is divided by the point of contact }T,
\displaystyle \text{are of lengths }14\text{ cm and }16\text{ cm respectively. If area of }\triangle PQR\text{ is }336\text{ cm}^2\text{,}
\displaystyle \text{find the sides }PQ\text{ and }PR\text{.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:} \displaystyle QR=QT+TR=14+16=30\text{ cm}.
\displaystyle \text{Let }s\text{ be the semi-perimeter of }\triangle PQR.
\displaystyle \text{Area of }\triangle PQR=r\times s
\displaystyle 336=8s
\displaystyle \therefore s=42\text{ cm}.
\displaystyle \text{Let the circle touch }PQ\text{ and }PR\text{ at }S\text{ and }U\text{ respectively.}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle QS=QT=14\text{ cm},\qquad RU=RT=16\text{ cm}
\displaystyle \text{and let }PS=PU=x\text{ cm}.
\displaystyle PQ=x+14\qquad\text{and}\qquad PR=x+16
\displaystyle s=\frac{PQ+QR+PR}{2}
\displaystyle 42=\frac{(x+14)+30+(x+16)}{2}
\displaystyle 84=2x+60
\displaystyle 2x=24
\displaystyle \therefore x=12\text{ cm}.
\displaystyle PQ=12+14=26\text{ cm}
\displaystyle PR=12+16=28\text{ cm}
\displaystyle \therefore PQ=26\text{ cm and }PR=28\text{ cm}.
\displaystyle \\

 

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