Solution

(b) Let $O$ be the centre of two concentric circles $C_1$ and $C_2$, whose radii are $r_1=4cm$ and $r_2=5cm$. Now, we draw a chord $AC$ of circle $C_2$, which touches the circle $C_1$ at $B$.

Also, join $OB$, which is perpendicular to $AC$. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

Now, in right angled $\mathrm{△}OBC$, by using Pythagoras theorem,

$O{C}^2=B{C}^2+B{O}^2$

$\Rightarrow 5^2=B{C}^2+4^2$

$[\because .$ hypotenuse ${.}^2=(\text{ base }{)}^2+(\text{ perpendicular }{)}^2]$

$\Rightarrow B{C}^2=25-16=9\Rightarrow BC=3cm$

$\therefore$ Length of chord $AC=2BC=2\times 3=6cm$

Solution

(d) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

i.e.,

$\begin{array}{r}\mathrm{\angle }AOB+\mathrm{\angle }COD={180}^{\circ }\\ \mathrm{\angle }COD={180}^{\circ }-\mathrm{\angle }AOB\\ & ={180}^{\circ }-{125}^{\circ }={55}^{\circ }\end{array}$

$\Rightarrow \mathrm{\angle }COD={180}^{\circ }-\mathrm{\angle }AOB$

Solution

(c) In figure, $AOC$ is a diameter of the circle. We know that, diameter subtends an angle ${90}^{\circ }$ at the circle.

So,

$\begin{array}{r}\mathrm{\angle }ABC={90}^{\circ }\\ \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }\end{array}$

In $\mathrm{△}ACB$,

$\Rightarrow$[since, sum of all angles of a triangle is ${180}^{\circ }$ ]

$\Rightarrow$ $\mathrm{\angle }A+{90}^{\circ }+{50}^{\circ }={180}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }A+140=180$

$\begin{array}{r}\mathrm{\angle }A={180}^{\circ }-{140}^{\circ }={40}^{\circ }\\ \mathrm{\angle }A\text{ or }\mathrm{\angle }OAB={40}^{\circ }\end{array}$

Now, $AT$ is the tangent to the circle at point $A$. So, $OA$ is perpendicular to $AT$.

$\begin{array}{rl}& \therefore \mathrm{\angle }OAT={90}^{\circ }\text{ [from figure] }\\ & \Rightarrow \mathrm{\angle }OAB+\mathrm{\angle }BAT={90}^{\circ }\\ & \text{ On putting }\mathrm{\angle }OAB={40}^{\circ }\text{, we get }\\ & \Rightarrow \mathrm{\angle }BAT={90}^{\circ }-{40}^{\circ }={50}^{\circ }\end{array}$

Solution

(a) Firstly, draw a circle of radius $5cm$ having centre $O.P$ is a point at a distance of $13cm$ from $O$. A pair of tangents $PQ$ and $PR$ are drawn.

Thus, quadrilateral $PQOR$ is formed.

$\begin{array}{cccc}\because & OQ& \perp QP& \text{ [since, }AP\text{ is a tangent line] }\\ \text{ In right angled }\mathrm{△}PQO,& O{P}^2=O{Q}^2+Q{P}^2\\ \Rightarrow & {13}^2=5^2+Q{P}^2\\ \Rightarrow & Q{P}^2=169-25=144\\ \Rightarrow & QP=12cm\\ & & \\ \text{ Now, }& ={\displaystyle \frac{1}{2}}\times 12\times 5=30c{m}^2\\ \therefore & \text{ area of }\mathrm{△}OQP={\displaystyle \frac{1}{2}}\times QP\times QO\\ & & \end{array}$

Solution

(d) First, draw a circle of radius $5cm$ having centre $O$. A tangent $XY$ is drawn at point $A$.

A chord $CD$ is drawn which is parallel to $XY$ and at a distance of $8cm$ from $A$. Now, $\mathrm{\angle }OAY={90}^{\circ }$

[Tangent and any point of a circle is perpendicular to the radius through the point of contact] $\Rightarrow$ $\mathrm{△}OAY+\mathrm{△}OED={180}^{\circ }$

Also,

$\mathrm{△}OED={180}^{\circ }$

Now, in right angled $\mathrm{△}OEC$,

$AE=8cm.\text{ Join }OC$

$\Rightarrow$ $O{C}^2=O{E}^2+E{C}^2$

$\begin{array}{cc}E{C}^2=O{C}^2-O{E}^2& [\text{ by Pythagoras theorem }]\\ & =5^2-3^2& \end{array}$

$[\because OC=$ radius $=5cm,OE=AE-AO=8-5=3cm]$ $=25-9=16$

$\Rightarrow$ $EC=4cm$

Hence,

length of chord $CD=2CE=2\times 4=8cm$

[since, perpendicular from centre to the chord bisects the chord]

Solution

(c) Join $OA$.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore$ $\mathrm{\angle }OAT$ $={90}^{\circ }$

In $\mathrm{△}OAT$, $\cos {30}^{\circ }$ $={\displaystyle \frac{AT}{OT}}$

$\Rightarrow$ ${\displaystyle \frac{\sqrt{3}}{2}}$ $={\displaystyle \frac{AT}{4}}$

$\Rightarrow$ $AT$ $=2\sqrt{3}cm$

Solution

(a) Given, $\mathrm{\angle }QPR={50}^{\circ }$

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore$ $\mathrm{\angle }OPR={90}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }OPQ+\mathrm{\angle }QPR={90}^{\circ }$

[from figure]

$\Rightarrow$ $\mathrm{\angle }OPQ={90}^{\circ }-{50}^{\circ }={40}^{\circ }$

$[\because \mathrm{\angle }QPR={50}^{\circ }]$

Now,

$OP=OQ=$ Radius of circle

$\therefore \mathrm{\angle }OQP=\mathrm{\angle }OPQ={40}^{\circ }$

In $\mathrm{△}OPQ$

[since, angles opposite to equal sides are equal]

[since, sum of angles of a triangle $={180}^{\circ }$ ]

$\Rightarrow$ $\mathrm{\angle }O+\mathrm{\angle }P+\mathrm{\angle }Q={180}^{\circ }$

$.+{40}^{\circ })[\because \mathrm{\angle }P={40}^{\circ }=\mathrm{\angle }Q]$

$\begin{array}{r}\mathrm{\angle }O={180}^{\circ }-({40}^{\circ }+{40}^{\circ })\\ & ={180}^{\circ }-{80}^{\circ }={100}^{\circ }\end{array}$

Solution

(a) Given, $PA$ and $PB$ are tangent lines.

[since, tangent at any point of a circle is perpendicular to the radius through the point of contact]

$\begin{array}{cc}\therefore & \mathrm{\angle }PAO={90}^{\circ }\\ \Rightarrow & \mathrm{\angle }PAB+\mathrm{\angle }BAO={90}^{\circ }\\ \Rightarrow & {65}^{\circ }+\mathrm{\angle }BAO={90}^{\circ }\\ \Rightarrow & \mathrm{\angle }BAO={90}^{\circ }-{65}^{\circ }={25}^{\circ }\end{array}$

Solution

(d) Let $P$ be an external point and a pair of tangents is drawn from point $P$ and angle between these two tangents is ${60}^{\circ }$.

Join $OA$ and $OP$.

Also, $OP$ is a bisector line of $\mathrm{\angle }APC$.

$\begin{array}{cc}\therefore & \mathrm{\angle }APO=\mathrm{\angle }CPO={30}^{\circ }\\ \text{ Also, }& OA\perp AP\end{array}$

Tangent at any point of a circle is perpendicular to the radius through the point of contact.

In right angled $\mathrm{△}OAP$,

$\begin{array}{r}\tan {30}^{\circ }={\displaystyle \frac{OA}{AP}}={\displaystyle \frac{3}{AP}}\\ {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{3}{AP}}\\ AP=3\sqrt{3}cm\end{array}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{3}{AP}}$

$\Rightarrow$ Hence, the length of each tangent is $3\sqrt{3}cm$.

Solution

(b) Given, $AB|PR$

Solution

False

Since a chord $AB$ subtends an angle of ${60}^{\circ }$ at the centre of a circle.

i.e.,

$\mathrm{\angle }AOB={60}^{\circ }$

As $OA=OB=$ Radius of the circle

$\therefore \mathrm{\angle }OAB=\mathrm{\angle }OBA={60}^{\circ }$

The tangent at points $A$ and $B$ is drawn, which intersect at $C$.

We know, $OA\perp AC$ and $OB\perp BC$.

$\therefore \mathrm{\angle }OAC={90}^{\circ },\mathrm{\angle }OBC={90}^{\circ }$

$\Rightarrow \mathrm{\angle }OAB+\mathrm{\angle }BAC={90}^{\circ }$

and $\mathrm{\angle }OBA+\mathrm{\angle }ABC={90}^{\circ }$

$\Rightarrow \mathrm{\angle }BAC={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

and $\mathrm{\angle }ABC={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

In $\mathrm{△}ABC,\mathrm{\angle }BAC+\mathrm{\angle }CBA+\mathrm{\angle }ACB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$

Solution

False

Because the length of tangent from an external point $P$ on a circle may or may not be greater than the radius of the circle.

Solution

True

$PT$ is a tangent drawn from external point $P$. Join $OT$.

$\because OT\perp PT$

So, OPT is a right angled triangle formed.

In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle. $\therefore$ or $OP>PT$ $PT<OP$

Solution

True

This may be possible only when both tangent lines coincide or are parallel to each other.

Solution

True

From point $P$, two tangents are drawn.

Given,

$OT=a$

Also, line $OP$ bisects the $\mathrm{\angle }RPT$.

$\therefore$ Also,

In right angled $\mathrm{△}OTP$,

$\mathrm{\angle }TPO=\mathrm{\angle }RPO={45}^{\circ }$

$OT\perp PT$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{a}{OP}}\Rightarrow OP=a\sqrt{2}$

Solution

False

From point $P$, two tangents are drawn.

Given,

$OT=a$

Also, line $OP$ bisects the $\mathrm{\angle }RPT$.

$\begin{array}{cc}\therefore & \mathrm{\angle }TPO=\mathrm{\angle }RPO={30}^{\circ }\\ \text{ Also, }& OT\perp PT\end{array}$

In right angled $\mathrm{△}OTP$,

$\begin{array}{cc}& \sin {30}^{\circ }={\displaystyle \frac{OT}{OP}}\\ \Rightarrow & {\displaystyle \frac{1}{2}}={\displaystyle \frac{a}{OP}}\\ \Rightarrow & OP=2a\end{array}$

Solution

True

Let $EAF$ be tangent to the circumcircle of $\mathrm{△}ABC$.

To prove

Here,

$\Rightarrow$ $EAF|BC$ $\mathrm{\angle }EAB=\mathrm{\angle }ABC$

$AB=AC$

$\mathrm{\angle }ABC=\mathrm{\angle }ACB$

[angle between tangent and is chord equal to angle made by chord in the alternate segment] $\therefore$ Also, $\mathrm{\angle }EAB=\mathrm{\angle }BCA$

From Eqs. (i) and (ii), we get

$\Rightarrow EAF|BC$

Solution

False

Given that $PQ$ is any line segment and $S_1,S_2,S_3,S_4,\dots$ circles are touch a line segment $PQ$ at a point $A$. Let the centres of the circles $S_1,S_2,S_3,S_4,\dots$ be $C_1,C_2,C_3,C_4,\dots$ respectively.

To prove Centres of these circles lie on the perpendicular bisector of $PQ$.

Now, joining each centre of the circles to the point $A$ on the line segment $PQ$ by a line segment i.e., $C_{1}A,C_{2}A,C_{3}A,C_{4}A$,… so on.

We know that, if we draw a line from the centre of a circle to its tangent line, then the line is always perpendicular to the tangent line. But it not bisect the line segment $PQ$.

So,

Since, each circle is passing through a point $A$. Therefore, all the line segments $C_{1}A,C_{2}A,C_{3}A,C_{4}A,\dots$, so on are coincident.

So, centre of each circle lies on the perpendicular line of $PQ$ but they do not lie on the perpendicular bisector of $PQ$.

Hence, a number of circles touch a given line segment $PQ$ at a point $A$, then their centres lie

Solution

True

We draw two circle with centres $C_1$ and $C_2$ passing through the end points $P$ and $Q$ of a line segment $PQ$. We know, that perpendicular bisectors of a chord of a circle always passes through the centre of circle.

Thus, perpendicular bisector of $PQ$ passes through $C_1$ and $C_2$. Similarly, all the circle passing through $PQ$ will have their centre on perpendiculars bisectors of $PQ$.

Solution

True

To Prove, $BC=BD$

Join $BC$ and $OC$.

Given,

$\Rightarrow$ $\mathrm{\angle }BAC={30}^{\circ }$

$\mathrm{\angle }BCD={30}^{\circ }$

[angle between tangent and chord is equal to angle made by chord in the alternate

$\begin{array}{rl}& \therefore \mathrm{\angle }ACD=\mathrm{\angle }ACO+\mathrm{\angle }OCD={30}^{\circ }+{90}^{\circ }={120}^{\circ }\\ & [\because OC\perp CD\text{ and }OA=OC=\text{ radius }\Rightarrow \mathrm{\angle }OAC=\mathrm{\angle }OCA={30}^{\circ }]\\ & \text{ In }\mathrm{△}ACD\\ & \mathrm{\angle }CAD+\mathrm{\angle }ACD+\mathrm{\angle }ADC={180}^{\circ }\\ & \Rightarrow {30}^{\circ }+{120}^{\circ }+\mathrm{\angle }ADC={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }ADC={180}^{\circ }-({30}^{\circ }+{120}^{\circ })={30}^{\circ }\\ & \text{ Now, in }\mathrm{△}BCD\\ & \mathrm{\angle }BCD=\mathrm{\angle }BDC={30}^{\circ }\\ & BC=BD\end{array}$

[since, sides opposite to equal angles are equal]

Solution

Let $C_1$ and $C_2$ be the two circles having same centre $O$. $AC$ is a chord which touches the $C_1$ at point $D$.

Solution

Given Two tangents $PQ$ and $PR$ are drawn from an external point to a circle with centre $O$.

To prove $QORP$ is a cyclic quadrilateral. proof Since, $PR$ and $PQ$ are tangents.

So,

$OR\perp PR$ and $OQ\perp PQ$

[since, if we drawn a line from centre of a circle to its tangent line. Then, the line always

$\begin{array}{cc}\therefore & \mathrm{\angle }ORP=\mathrm{\angle }OQP={90}^{\circ }\\ \text{ Hence, }& \mathrm{\angle }ORP+\mathrm{\angle }OQP={180}^{\circ }\end{array}$

perpendicular to the tangent line]

So, QOPR is cyclic quadrilateral.

[If sum of opposite angles is quadrilateral in ${180}^{\circ }$, then the quadrilateral is cyclic]

Hence proved.

Solution

Given Two tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with centre $O$.

To prove Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

In $\mathrm{\angle }RPQ$.

Construction Join $OR$, and $OQ$.

In $\mathrm{△}POR$ and $\mathrm{△}POQ$

$\mathrm{\angle }PRO=\mathrm{\angle }PQO={90}^{\circ }$

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

$OR=OQ$

[radii of some circle]

Since, $OP$ is common. $\therefore$ $\mathrm{△}PRO\cong \mathrm{△}PQO$ Hence, $\mathrm{\angle }RPO=\mathrm{\angle }QPO$

[RHS]

[by $CPCT]$

Thus, $O$ lies on angle bisecter of $PR$ and $PQ$.

Hence proved.

Solution

Two tangents $BD$ and $BC$ are drawn from an external point $B$.

To prove

Given,

Join $OC,OD$ and $BO$.

Since, $BC$ and $BD$ are tangents.

$\therefore OC\perp BC$ and $OD\perp BD$

We know, $OB$ is a angle bisector of $\mathrm{\angle }DBC$.

$\therefore \mathrm{\angle }OBC=\mathrm{\angle }DBO={60}^{\circ }$

In right angled $\mathrm{△}OBC$,

$\cos {60}^{\circ }={\displaystyle \frac{BC}{OB}}$

$\Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{BC}{OB}}$

$\Rightarrow OB=2BC$

Also, $BC=BD$

[tangent drawn from internal point to circle are equal]

$\therefore OB=BC+BC$

$\Rightarrow OB=BC+BD$

Solution

Given $AB$ and $CD$ are common tangent to two circles of unequal radius To prove $AB=CD$

Construction Produce $AB$ and $CD$, to intersect at $P$.

Proof $PA=PC$

[the length of tangents drawn from an internal point to a circle are equal]

Also,

$PB=PD$

[the lengths of tangents drawn from an internal point to a circle are equal]

$\therefore PA-PB=PC-PD$

$AB=CD$

Hence proved.

Solution

Given $AB$ and $CD$ are tangents to two circles of equal radii. To prove

$AB=CD$

Construction Join $OA,OC,O^{\prime }B$ and $O^{\prime }D$

Proof Now, $\mathrm{\angle }OAB={90}^{\circ }$

[tangent at any point of a circle is perpendicular to radius through the point of contact] Thus, $AC$ is a straight line.

Also,

$\therefore$

Similarly, $BD$ is a straight line and

Also,

In quadrilateral $ABCD$, and $ABCD$ is a rectangle Hence,

$\begin{array}{c}\mathrm{\angle }OAB+\mathrm{\angle }OCD={180}^{\circ }\\ AB|CD\end{array}$

$\begin{array}{r}\mathrm{\angle }{O}^{\prime }BA=\mathrm{\angle }{O}^{\prime }DC={90}^{\circ }\\ AC=BD\text{ [radii of two circles are equal] }\\ \mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }\\ AC=BD\end{array}$

$AB=CD$ [opposite sides of rectangle are equal]

Solution

Given Common tangents $AB$ and $CD$ to two circles intersecting at $E$.

To prove $A$

$B=CD$

Proof

$$\begin{array}{}\text{(i)}& EA=EC\end{array}$$

[the lengths of tangents drawn from an internal point to a circle are equal]

$$\begin{array}{}\text{(ii)}& EB=ED\end{array}$$

On adding Eqs. (i) and (ii), we get

$\begin{array}{r}EA+EB=EC+ED\\ AB=CD\end{array}$

Hence proved.

Solution

Given Chord $PQ$ is parallel to tangent at $R$.

To prove $R$ bisects the arc $PRQ$

$\mathrm{\angle }1=\mathrm{\angle }3$

[angle between tangent and chord is equal to angle made by chord in alternate segment]

$\begin{array}{cc}\therefore & \mathrm{\angle }2=\mathrm{\angle }3\\ \Rightarrow & PR=QR\\ \Rightarrow & PR=QR\end{array}$

Solution

To prove $\mathrm{\angle }1=\mathrm{\angle }2$, let $PQ$ be a chord of the circle. Tangents are drawn at the points $R$ and Q.

Let $P$ be another point on the circle, then, join $PQ$ and $PR$.

Since, at point $Q$, there is a tangent.

$\therefore \mathrm{\angle }2=\mathrm{\angle }P$ [angles in alternate segments are equal]

Since, at point $R$, there is a tangent.

$\therefore \mathrm{\angle }1=\mathrm{\angle }P$ [angles in alternate segments are equal]

$\therefore \mathrm{\angle }1=\mathrm{\angle }2=\mathrm{\angle }P$

Hence proved.

Solution

Given, $AB$ is a diameter of the circle.

A tangent is drawn from point $A$. Draw a chord $CD$ parallel to the tangent MAN.

So, $CD$ is a chord of the circle and $OA$ is a radius of the circle.

$\mathrm{\angle }MAO={90}^{\circ }$

[tangent at any point of a circle is perpendicular to the radius through the point of contact] $\mathrm{\angle }CEO=\mathrm{\angle }MAO$ $\mathrm{\angle }CEO={90}^{\circ }$

[corresponding angles]

$\therefore$

Thus, $OE$ bisects $CD,$ [perpendicular from centre of circle to chord bisects the chord] Similarly, the diameter $AB$ bisects all. Chords which are parallel to the tangent at the point $A$.

Solution

Given $A$ hexagon $ABCDEF$ circumscribe a circle.

To prove $AB+CD+EF=BC+DE+FA$

Proof $AB+CD+EF=(AQ+QB)+(CS+SD)+(EU+UF)$

$=AP+BR+CR+DT+ET+FP$

$=(AP+FP)+(BR+CR)+(DT+ET)$

$\begin{array}{r}AB+CD+EF=AF+BC+DE\\ AQ=AP\\ QB=BR\\ CS=CR\\ DS=DT\\ EU=ET\end{array}$

[tangents drawn from an external point to a circle are equal]

Hence proved.

Solution

A circle is inscribed in the $\mathrm{△}ABC$, which touches the $BC,CA$ and $AB$.

Given,

$BC=a,CA=b\text{ and }AB=c$

By using the property, tangents are drawn from an external point to the circle are equal in length.

$\begin{array}{rl}& \therefore BD=BF=x\text{ [say] }\\ & \text{ ond }\\ & \text{ Now, }BC+CA+AB=a+b+c\\ & \Rightarrow (BD+DC)+(CE+EA)+(AF+FB)=a+b+c\\ & \Rightarrow (x+y)+(y+z)+(z+x)=a+b+c\\ & \Rightarrow 2(x+y+z)=2s\\ & \Rightarrow s=x+y+z\\ & \Rightarrow x=s-(y+z)\\ & \Rightarrow BD=s-b[\because b=AE+EC=z+y]\end{array}$

Solution

Two tangents $PA$ and $PB$ are drawn to a circle with centre $O$ from an external point $P$.

Solution

Since, $AC$ is a diameter line, so angle in semi-circle makes an angle ${90}^{\circ }$. $\therefore$ In $\mathrm{△}ABC$, $\mathrm{\angle }ABC={90}^{\circ }$ $\mathrm{\angle }CAB+\mathrm{\angle }ABC+\mathrm{\angle }ACB={180}^{\circ }$ [by property]

$[\because .$ sum of all interior angles of any triangle is ${.180}^{\circ }]$

$\Rightarrow \mathrm{\angle }CAB+\mathrm{\angle }ACB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

Since, diameter of a circle is perpendicular to the tangent.

i.e.

$\therefore$ $CA\perp AT$

$\mathrm{\angle }CAT={90}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }CAB+\mathrm{\angle }BAT={90}^{\circ }$

From Eqs. (i) and (ii),

$\begin{array}{rl}\Rightarrow & \mathrm{\angle }CAB+\mathrm{\angle }ACB=\mathrm{\angle }CAB+\mathrm{\angle }BAT\\ \mathrm{\angle }ACB=\mathrm{\angle }BAT\end{array}$

Hence proved.

Solution

Here, two circles are of radii $OP=3cm$ and $P{O}^{\prime }=4cm$.

These two circles intersect at $P$ and $Q$.

Here, $OP$ and $P{O}^{\prime }$ are two tangents drawn at point $P$.

$\mathrm{\angle }OP{O}^{\prime }={90}^{\circ }$

[tangent at any point of circle is perpendicular to radius through the point of contact]

Join $O{O}^{\prime }$ and $PN$.

In right angled $\mathrm{△}OP{O}^{\prime }$,

$\begin{array}{rl}& (O{O}^{\prime }{)}^2=(OP{)}^2+(P{O}^{\prime }{)}^2\text{ [by Pythagoras theorem] }\\ & \text{ i.e., }(\text{ Hypotenuse }{)}^2=(\text{ Base }{)}^2+(\text{ Perpendicular }{)}^2\\ & =(3{)}^2+(4{)}^2=25\\ & \Rightarrow O{O}^{\prime }=5cm\\ & \text{ Also, }PN\perp O{O}^{\prime }\end{array}$

$(OP{)}^2=(ON{)}^2+(NP{)}^2[\text{ by Pythagoras theorem }]$

$$\begin{array}{}\text{(i)}& \Rightarrow (NP{)}^2=3^2-x^2=9-x^2\end{array}$$

and in right angled $\mathrm{△}PN{O}^{\prime }$,

$\begin{array}{rl}& (P{O}^{\prime }{)}^2=(PN{)}^2+(N{O}^{\prime }{)}^2[\text{ by Pythagoras theorem }]\\ & \Rightarrow (4{)}^2=(PN{)}^2+(5-x{)}^2\\ & \Rightarrow (PN{)}^2=16-(5-x{)}^2\end{array}$

From Eqs. (i) and (ii),

$9-x^2=16-(5-x{)}^2$

$\Rightarrow 7+x^2-(25+x^2-10x)=0$

$\Rightarrow 10x=18$

$\therefore x=1.8$

Again, in right angled $\mathrm{△}OPN$,

$\begin{array}{cc}\Rightarrow & 3^2=(1.8{)}^2+(NP{)}^2\\ \Rightarrow & (NP{)}^2=9-3.24=5.76\\ \therefore & (NP)=2.4\\ \therefore \text{ Length of common chord, }& PQ=2PN=2\times 2.4=4.8cm\end{array}$

[by Pythagoras theorem]

Solution

Let $O$ be the centre of the given circle. Suppose, the tangent at $P$ meets $BC$ at $Q$. Join $BP$.

Proof

[tangent at any point of circle is perpendicular to radius through the point of contact]

$\therefore \ln \mathrm{△}ABC,\begin{array}{r}\mathrm{\angle }1+\mathrm{\angle }5={90}^{\circ }\\ \mathrm{\angle }3=\mathrm{\angle }1\end{array}$ [angle sum property, $\mathrm{\angle }ABC={90}^{\circ }$ ]

[angle between tangent and the chord equals angle made by the chord in alternate segment]

$\therefore$ $\mathrm{\angle }3+\mathrm{\angle }5={90}^{\circ }$ $\dots$ (i)

Also, $\mathrm{\angle }APB={90}^{\circ }$ [angle in semi-circle]

$\Rightarrow$ $\mathrm{\angle }3+\mathrm{\angle }4={90}^{\circ }$ $[\mathrm{\angle }APB+\mathrm{\angle }BPC={180}^{\circ }.$, linear pair]

From Eqs. (i) and (ii), we get

$\Rightarrow$ $\mathrm{\angle }3+\mathrm{\angle }5=\mathrm{\angle }3+\mathrm{\angle }4$

$\Rightarrow$ $\mathrm{\angle }5=\mathrm{\angle }4$

Also, $PQ=QC$ [sides opposite to equal angles are equal]

$\Rightarrow$ $QP=QB$

$\Rightarrow$ $[$ tangents drawn from an internal point to a circle are equal]

$QB=QC$

Solution

$PQ$ and $PR$ are two tangents drawn from an external point $P$.

$\therefore PQ=PR$

[the lengths of tangents drawn from an external point to a circle are equal]

$\Rightarrow$ $\mathrm{\angle }PQR=\mathrm{\angle }QRP$

[angles opposite to equal sides are equal]

Now, in $\mathrm{△}PQR\mathrm{\angle }PQR+\mathrm{\angle }QRP+\mathrm{\angle }RPQ={180}^{\circ }$

[sum of all interior angles of any triangle is ${180}^{\circ }$ ]

$\Rightarrow \mathrm{\angle }PQR+\mathrm{\angle }PQR+{30}^{\circ }={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }PQR={180}^{\circ }-{30}^{\circ }$

$\Rightarrow \mathrm{\angle }PQR={\displaystyle \frac{{180}^{\circ }-{30}^{\circ }}{2}}={75}^{\circ }$

Since, $SR|QP$

$\therefore \mathrm{\angle }SRQ=\mathrm{\angle }RQP={75}^{\circ }$ [alternate interior angles]

Also, $\mathrm{\angle }PQR=\mathrm{\angle }QSR={75}^{\circ }$ [by alternate segment theorem]

In $\mathrm{△}QRS,\mathrm{\angle }Q+\mathrm{\angle }R+\mathrm{\angle }S={180}^{\circ }$

[sum of all interior angles of any triangle is ${180}^{\circ }$ ]

$\begin{array}{cc}\Rightarrow & \mathrm{\angle }Q={180}^{\circ }-({75}^{\circ }+{75}^{\circ })\\ & ={30}^{\circ }\\ \therefore & \mathrm{\angle }RQS={30}^{\circ }\end{array}$

Solution

A circle is drawn with centre $O$ and $AB$ is a diameter.

$AC$ is a chord such that $\mathrm{\angle }BAC={30}^{\circ }$.

Given $AB$ is $\alpha$ diameter and $AC$ is a chord of circle with certre $O,\mathrm{\angle }BAC={30}^{\circ }$.

To prove

$\ln \mathrm{△}ABC$,

$\begin{array}{rl}& \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }\\ & {30}^{\circ }+\mathrm{\angle }B+{90}^{\circ }={180}^{\circ }\\ & \Rightarrow \\ & \text{ Also, }\\ & \Rightarrow \\ & \mathrm{\angle }CBA+\mathrm{\angle }CBD={180}^{\circ }\\ & \mathrm{\angle }CBD={180}^{\circ }-{60}^{\circ }-{120}^{\circ }\\ & [\because \mathrm{\angle }CBA={60}^{\circ }]\\ & \Rightarrow {120}^{\circ }+\mathrm{\angle }BDC+{30}^{\circ }={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }BDC={30}^{\circ }\\ & \text{ From Eqs. (i) and (ii), }\\ & \mathrm{\angle }BCD=\mathrm{\angle }BDC\\ & BC=BD\end{array}$

Solution

Let mid-point of an $arcAMB$ be $M$ and $TM{T}^{\prime }$ be the tangent to the circle.

Join $AB,AM$ and $MB$.

Since,

$\begin{array}{r}arcAM=arcMB\\ \text{ Chord }AM=\text{ Chord }MB\\ AM=MB\\ \mathrm{\angle }MAB=\mathrm{\angle }MBA\end{array}$

$\Rightarrow$ In $\mathrm{△}AMB$,

$\Rightarrow$ [equal sides corresponding to the equal angle] …(i)

Since, $TM{T}^{\prime }$ is a tangent line. $\therefore$ $\mathrm{\angle }AMT=\mathrm{\angle }MBA$ [angles in alternate segments are equal] $=\mathrm{\angle }MAB$ [from Eq. (i)]

But $\mathrm{\angle }AMT$ and $\mathrm{\angle }MAB$ are alternate angles, which is possible only when

$AB|TM{T}^{\prime }$

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Hence proved.

Solution

Joint $AO,OC$ and $O^{\prime }D,O^{\prime }B$. Now, in $\mathrm{△}E{O}^{\prime }D$ and $\mathrm{△}E{O}^{\prime }B$,

$$\begin{array}{r}\text{(radius)}& O^{\prime }D=O^{\prime }B\text{(commonside)}& O^{\prime }E=O^{\prime }EED=EB\end{array}$$

[since, tangents drawn from an external point to the circle are equal in length]

$\begin{array}{rl}& \therefore \mathrm{△}E{O}^{\prime }D\cong \mathrm{△}E{O}^{\prime }B\text{ [by SSS congruence rule] }\\ & \Rightarrow \mathrm{\angle }{O}^{\prime }ED=\mathrm{\angle }{O}^{\prime }EB\end{array}$