Solution

Let $O$ be the centre of the circle.

Given that,

$OQ=25 cm$ and $PQ=24 cm$

As the radius is perpendicular to the tangent at the point of contact,

Therefore, $OP \perp PQ$

Applying Pythagoras theorem in $\triangle OPQ$, we obtain

$OP^{2}+PQ^{2}=OQ^{2}$

$OP^{2}+24^{2}=25^{2}$

$OP^{2}=625-576$

$OP^{2}=49$

$OP=7$

Therefore, the radius of the circle is $7 cm$.

Hence, alternative (A) is correct.

Solution

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OP \perp TP$ and $OQ \perp TQ$

$\angle OPT=90^{\circ}$

$\angle OQT=90^{\circ}$

In quadrilateral POQT,

Sum of all interior angles $=360^{\circ}$

$\angle OPT+\angle POQ+\angle OQT+\angle PTQ=360^{\circ}$

$\Rightarrow 90^{\circ}+110^{\circ}+90^{\circ}+\angle PTQ=360^{\circ}$

$\Rightarrow \angle PTQ=70^{\circ}$

Hence, alternative (B) is correct.

Solution

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OA \perp PA$ and $OB \perp PB$

$\angle OBP=90^{\circ}$

$\angle OAP=90^{\circ}$

In AOBP,

Sum of all interior angles $=360^{\circ}$

$ \begin{aligned} & \angle OAP+\angle APB+\angle PBO+\angle BOA=360^{\circ} \\ & 90^{\circ}+80^{\circ}+90^{\circ}+\angle BOA=360^{\circ} \\ & \angle BOA=100^{\circ} \end{aligned} $

In $\triangle OPB$ and $\triangle OPA$,

$A P=B P($ Tangents from a point $)$

$OA=OB$ (Radii of the circle)

$OP=OP$ (Common side)

Therefore, $\triangle OPB \cong \triangle OPA$ (SSS congruence criterion)

$A \rightarrow B, P \rightarrow P, O \rightarrow O$

And thus, $\angle POB=\angle POA$

$\angle POA=\dfrac{1}{2} \angle AOB=\dfrac{100^{\circ}}{2}=50^{\circ}$

Hence, alternative (A) is correct.

Solution

Let $A B$ be a diameter of the circle. Two tangents $P Q$ and $R S$ are drawn at points $A$ and $B$ respectively. Radius drawn to these tangents will be perpendicular to the tangents.

Thus, $OA \perp RS$ and $OB \perp PQ$

$\angle OAR=90^{\circ}$

$\angle OAS=90^{\circ}$

$\angle OBP=90^{\circ}$

$\angle OBQ=90^{\circ}$

It can be observed that $\angle OAR=\angle OBQ$ (Alternate interior angles)

$\angle OAS=\angle OBP$ (Alternate interior angles)

Since alternate interior angles are equal, lines $P Q$ and $R S$ will be parallel.

Solution

Let us consider a circle with centre $O$. Let $AB$ be a tangent which touches the circle at $P$.

We have to prove that the line perpendicular to $A B$ at $P$ passes through centre $O$. We shall prove this by contradiction method.

Let us assume that the perpendicular to $A B$ at $P$ does not pass through centre $O$. Let it pass through another point $O^{\prime}$. Join OP and O’P.

As perpendicular to $A B$ at $P$ passes through $O^{\prime}$, therefore,

$\angle O^{\prime} PB=90^{\circ}$

$O$ is the centre of the circle and $P$ is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

$\therefore \angle OPB=90^{\circ}$..

Comparing equations (1) and (2), we obtain

$\angle O ’ PB=\angle OPB$..

From the figure, it can be observed that,

$\angle O$ ‘PB $<\angle OPB$…

Therefore, $\angle O O^{\prime} PB=\angle OPB$ is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to $A B$ through $P$ passes through centre $O$.

Solution

Let us consider a circle centered at point $O$.

$A B$ is a tangent drawn on this circle from point $A$.

Given that,

$O A=5 cm$ and $A B=4 cm$

In $\triangle ABO$,

$OB \perp AB$ (Radius $\perp$ tangent at the point of contact)

Applying Pythagoras theorem in $\triangle A B O$, we obtain

$AB^{2}+BO^{2}=OA^{2}$

$4^{2}+BO^{2}=5^{2}$

$16+BO^{2}=25$

$BO^{2}=9$

$BO=3$

Hence, the radius of the circle is $3 cm$.

Solution

Let the two concentric circles be centered at point $O$. And let $PQ$ be the chord of the larger circle which touches the smaller circle at point $A$. Therefore, $PQ$ is tangent to the smaller circle.

$OA \perp PQ(As OA$ is the radius of the circle)

Applying Pythagoras theorem in $\triangle OAP$, we obtain

$OA^{2}+AP^{2}=OP^{2}$

$3^{2}+AP^{2}=5^{2}$

$9+AP^{2}=25$

$AP^{2}=16$

$AP=4$

In $\triangle OPQ$,

Since $OA \perp PQ$,

$AP=AQ$ (Perpendicular from the center of the circle bisects the chord)

$\therefore PQ=2 AP=2 \times 4=8$

Therefore, the length of the chord of the larger circle is $8 cm$.

Solution

It can be observed that

DR $=$ DS (Tangents on the circle from point D) $CR=CQ$ (Tangents on the circle from point C) …

$B P=B Q($ Tangents on the circle from point $B)$

AP $=$ AS (Tangents on the circle from point A)

Adding all these equations, we obtain

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(D R+C R)+(B P+A P)=(D S+A S)+(C Q+B Q)$

$C D+A B=A D+B C$

Solution

Let us join point $O$ to $C$.

In $\triangle OPA$ and $\triangle OCA$,

$OP=OC$ (Radii of the same circle)

$A P=A C($ Tangents from point $A)$

$AO=AO$ (Common side)

$\triangle OPA \cong \triangle OCA$ (SSS congruence criterion)

Therefore, $P \rightarrow C, A \rightarrow A, O \rightarrow O$

$\angle POA=\angle COA \ldots$ (i)

Similarly, $\triangle OQB \cong \triangle OCB$ $\angle QOB=\angle COB \ldots$

Since POQ is a diameter of the circle, it is a straight line.

Therefore, $\angle POA+\angle COA+\angle COB+\angle QOB=180^{\circ}$

From equations (i) and (ii), it can be observed that

$2 \angle COA+2 \angle COB=180^{\circ}$

$\angle COA+\angle COB=90^{\circ}$

$\angle A O B=90^{\circ}$

Solution

Let us consider a circle centered at point $O$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle which are touching the circle at point $A$ and $B$ respectively and $A B$ is the line segment, joining point of contacts $A$ and $B$ together such that it subtends $\angle A O B$ at center $O$ of the circle.

It can be observed that

$OA$ (radius) $\perp PA$ (tangent)

Therefore, $\angle OAP=90^{\circ}$

Similarly, OB (radius) $\perp$ PB (tangent)

$\angle OBP=90^{\circ}$

In quadrilateral OAPB,

Sum of all interior angles $=360^{\circ}$

$\angle OAP+\angle APB+\angle PBO+\angle BOA=360^{\circ}$

$90^{\circ}+\angle APB+90^{\circ}+\angle BOA=360^{\circ}$

$\angle APB+\angle BOA=180^{\circ}$

Hence, it can be observed 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.

Solution

Since $A B C D$ is a parallelogram,

$A B=C D$…(1)

$B C=A D \ldots(2)$

It can be observed that

DR $=$ DS (Tangents on the circle from point D)

$CR=CQ$ (Tangents on the circle from point C)

$BP=BQ$ (Tangents on the circle from point $B$ )

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(DS+AS)+(CQ+BQ)$

$C D+A B=A D+B C$

On putting the values of equations (1) and (2) in this equation, we obtain

$2 AB=2 BC$

$AB=BC$

Comparing equations (1), (2), and (3), we obtain

$AB=BC=CD=DA$

Hence, $A B C D$ is a rhombus.

Solution

Let the given circle touch the sides $A B$ and $A C$ of the triangle at point $E$ and $F$ respectively and the length of the line segment AF be $x$.

In $\triangle ABC$,

$CF=CD=6 cm$ (Tangents on the circle from point $C$ )

$BE=BD=8 cm$ (Tangents on the circle from point $B$ )

$AE=AF=x($ Tangents on the circle from point $A)$

$AB=AE+EB=x+8$

$BC=BD+DC=8+6=14$

$CA=CF+FA=6+x$

$2 s=A B+B C+C A$

$=x+8+14+6+x$

$=28+2 x$

$s=14+x$

$ \text{ Area of } \begin{aligned} \Delta ABC & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{{14+x}{(14+x)-14}{(14+x)-(6+x)}{(14+x)-(8+x)}} \\ & =\sqrt{(14+x)(x)(8)(6)} \\ & =4 \sqrt{3(14 x+x^{2})} \end{aligned} $

Area of $\triangle OBC=\dfrac{1}{2} \times OD \times BC=\dfrac{1}{2} \times 4 \times 14=28$

Area of $\triangle OCA=\dfrac{1}{2} \times OF \times AC=\dfrac{1}{2} \times 4 \times(6+x)=12+2 x$

Area of $\triangle OAB=\dfrac{1}{2} \times OE \times AB=\dfrac{1}{2} \times 4 \times(8+x)=16+2 x$

Area of $\triangle ABC=$ Area of $\triangle OBC+$ Area of $\triangle OCA+$ Area of $\triangle OAB$

$4 \sqrt{3(14 x+x^{2})}=28+12+2 x+16+2 x$

$\Rightarrow 4 \sqrt{3(14 x+x^{2})}=56+4 x$

$\Rightarrow \sqrt{3(14 x+x^{2})}=14+x$

$\Rightarrow 3(14 x+x^{2})=(14+x)^{2}$

$\Rightarrow 42 x+3 x^{2}=196+x^{2}+28 x$

$\Rightarrow 2 x^{2}+14 x-196=0$

$\Rightarrow x^{2}+7 x-98=0$

$\Rightarrow x^{2}+14 x-7 x-98=0$

$\Rightarrow x(x+14)-7(x+14)=0$

$\Rightarrow(x+14)(x-7)=0$

Either $x+14=0$ or $x-7=0$

Therefore, $x=-14$ and 7

However, $x=-14$ is not possible as the length of the sides will be negative.

Therefore, $x=7$

Hence, $AB=x+8=7+8=15 cm$

$CA=6+x=6+7=13 cm$

Solution

Let $A B C D$ be a quadrilateral circumscribing a circle centered at $O$ such that it touches the circle at point $P, Q, R, S$. Let us join the vertices of the quadrilateral $A B C D$ to the center of the circle.

Consider $\triangle OAP$ and $\triangle OAS$,

AP $=$ AS (Tangents from the same point)

$OP=OS$ (Radii of the same circle)

$OA=OA$ (Common side)

$\Delta OAP \cong \triangle OAS$ (SSS congruence criterion)

And thus, $\angle POA=\angle AOS$

$\angle 1=\angle 8$

Similarly,

$\angle 2=\angle 3$

$\angle 4=\angle 5$

$\angle 6=\angle 7$

$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}$

$(\angle 1+\angle 8)+(\angle 2+\angle 3)+(\angle 4+\angle 5)+(\angle 6+\angle 7)=360^{\circ}$

$2 \angle 1+2 \angle 2+2 \angle 5+2 \angle 6=360^{\circ}$

$2(\angle 1+\angle 2)+2(\angle 5+\angle 6)=360^{\circ}$

$(\angle 1+\angle 2)+(\angle 5+\angle 6)=180^{\circ}$

$\angle AOB+\angle COD=180^{\circ}$

Similarly, we can prove that $\angle BOC+\angle DOA=180^{\circ}$

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.