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#9.4

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm.AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the $\bigtriangleup$ ABC.

Let us make a figure according to the question

Given : OA = 13 cm, Radius = 5 cm
Here AP = AQ            (tangent from the same point)
OP $\perp$ PA, OQ $\perp$ QA ( AP, AQ are tangents)
In $\bigtriangleup$ OPA using Pythagoras' theorem
$\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}$
$\left ( QP \right )^{2}= \left ( OP \right )^{2}+\left ( PA \right )^{2}$
$\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PA \right )^{2}$
$\left ( PA \right )^{2}= 169-25$
$PA= \sqrt{144}= 12\, cm$ ........(i)
Perimeter of $\bigtriangleup$ ABC = AB + BC + CA
= AB + BR + RC + CA
= AB + BD + CQ + CA
[ BP and BR are tangents from point B and CP and CQ are tangents from point C]
= AP + AQ      [ AP = AB + BP, AQ = AC + CQ]
= AP + AP      [ AP = AQ]
= 2AP
= 2 × 12           [using (i)]
= 24 cm

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#9.4

If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

According to question

In $\bigtriangleup ABD$ and $\bigtriangleup ACO$
AB = AC        [Given]
BO = CO        [Radius]
AO = AO        [Common side]
$\therefore \, \bigtriangleup ABO\cong \bigtriangleup ACO$ [By SSS congruence Criterion]
$\angle Q_{1}= \angle Q_{2}$ [CPCT]
In $\bigtriangleup ABD$ and $\bigtriangleup ACD$
AB = AC        [given]
$\angle Q_{1}= \angle Q_{2}$
AD = AD        [common side]
$\therefore \bigtriangleup ABD\cong \bigtriangleup ACD$    [By SAS congruence Criterion]
$\angle ADB= \angle ADC$ …..(i)   [CPCT]
$\angle ADB= \angle ADC= 180^{\circ}$ …..(ii)
From (i) and (ii)
$\angle ADB= 90^{\circ}$
OA is a perpendicular which bisects chord BC
Let AD = x, then OD = 9 – x      $\left ( \because OA= 9\, cm \right )$
Use Pythagoras in $\bigtriangleup$ ADC
$\left ( AC \right )^{2}= \left ( AD \right )^{2}+\left ( DC \right )^{2}$
$\left ( 6 \right )^{2}= x^{2}+\left ( DC \right )^{2}$
$\left ( AC \right )^{2}= 36-x^{2}$ …..(iii)
In $\bigtriangleup$ ODC using Pythagoras' theorem
$\left ( OC \right )^{2}= \left ( OD \right )^{2}+\left ( DC \right )^{2}$
$\left ( DC \right )^{2}= 81-\left ( 9-x \right )^{2}$ …..(iv)
From (iii) and (iv)
$36-x^{2}= 81-\left ( 9-x \right )^{2}$
$36-x^{2}- 81+\left ( 81+x^{2} -18x\right )= 0$
$\left [ \because \left ( a-b \right )^{2} +a^{2}+b^{2}-2ab\right ]$
$36-x^{2}-81+81+x^{2}-18x= 0$
$18x= 36$
$x= 2$
i.e., AD = 2 cm, OD = 9 – 2 = 7 cm
Put the value of x in (iii)
$\left ( DC \right )^{2}= 36-4$
$\left ( DC \right )^{2}= 32$
$DC= 4\sqrt{2}\, cm$
$BC= BD+DC$
$BC= 2DC$    $\left [ \because BD= DC \right ]$
$BC= 8\sqrt{2}\, cm$
Areas of $\bigtriangleup ABC= \frac{1}{2}\times base\times height$
$= \frac{1}{2}\times \times 8\sqrt{2}\times 2= 8\sqrt{2}\, cm$

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#9.4

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If $\angle$ PCA = $110^{\circ}$ , find $\angle$ CBA [see Figure].

Solution
Given : $\angle PCA= 110^{\circ}$
Here $\angle PCA= 90^{\circ}$ [ $\because$ PC is tangent]
$\angle PCA= \angle PCO+\angle OCA$
$110^{\circ}= 90^{\circ}+\angle OCA$
$\angle OCA= 20^{\circ}$
Here OC = OA (Radius)
$\therefore \: \angle OCA= \angle OAC= 20^{\circ}\cdots \left ( i \right )$ [ $\because$ Sides opposite to equal angles are equal]
We know that PC is tangent and angles in alternate segment are equal.
Hence $\angle BCP= \angle CAB= 20^{\circ}$
In $\bigtriangleup OAC$
$\angle O+\angle C+\angle A= 180^{\circ}$ [Interior angles sum of triangle is 180°]
$\angle O+20^{\circ}+20^{\circ}= 180^{\circ}$ [using (i)]
$\angle O= 180^{\circ}-40^{\circ}= 140^{\circ}$ …..(ii)
Here $\angle COB+\angle COA= 180^{\circ}$
$\angle COB= 180^{\circ}-140^{\circ}$ ( using (ii))
$\angle COB= 40^{\circ}$ …..(iii)
In $\bigtriangleup COB$
$\angle C+\angle O+\angle B= 180^{\circ}$ [using (iii)]
$90^{\circ}-20^{\circ}+40^{\circ}+\angle B= 180^{\circ}$
$\angle B= 180^{\circ}-110^{\circ}= 70^{\circ}$
Hence $\angle CBA= 70^{\circ}$

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#9.4

In Figure. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Given: Radius = 5 cm, OT = 13 cm
$OP\perp PT$ [ PT is tangent]
Using Pythagoras' theorem $\bigtriangleup$ OPT
$H^{2}+B^{2}+P^{2}$
$\left ( OT \right )^{2}= \left ( OP \right )^{2}+\left ( PT \right )^{2}$
$\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PT \right )^{2}$
$\left ( PT \right )^{2}= 169-25$
$PT= {\sqrt{144}}= 12\, cm$

PT and QT are tangents from the same point
$\therefore$ PT = QT = 12 cm
AT = PT – PA
AT = 12 – PA        ..…(i)
Similarly BT = 12 – QB          ..…(ii)
Since PA, PF and BF, BQ are tangents from points A and B respectively.
Hence, PA = AE              ..…(iii)
BQ = BE     ..…(iv)
AB is tangent at point E
Hence OE $\perp$ AB
$\angle AET= 180^{\circ}-\angle AEO$     $\left ( \because \, \angle AEO= 90^{\circ} \right )$
$\angle AET= 180^{\circ}-90^{\circ}$
$\angle AET= 90^{\circ}$
ET = OT – OF
ET = 13 – 5
ET = 8 cm
In $\bigtriangleup$ AET, using Pythagoras theorem
$\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}$
$\left ( AT \right )^{2}= \left ( AE \right )^{2}+\left ( ET \right )^{2}$
$\left ( 12-PA \right )^{2}= \left ( PA \right )^{2}+\left ( 8 \right )^{2}$   [using (i) and (ii)]
$144+\left ( PA \right )^{2}-24\left ( PA \right )-\left ( PA \right )^{2}-64= 0$
$\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]$
$80-42\left ( PA \right )= 0$
$24\left ( PA \right )= 80$
$\left ( PA \right )= \frac{80}{24}= \frac{10}{3}\, cm$
$\therefore \; AE= \frac{10}{3}\, cm\, \left [ using\left ( iii \right ) \right ]$
Similarly $BE= \frac{10}{3}\, cm$
$AB= AE+BE= \frac{10}{3}+\frac{10}{3}= \frac{20}{3}\, cm$

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#9.4

In Figure, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, and O' are collinear.

Construction: Join AO and OS
${O}'D$ and ${O}'B$

In $\bigtriangleup E{O}'B$ and $\bigtriangleup E{O}'D$
${O}'D= {O}'B$    [Radius are equal]
${O}'E= {O}'E$ [Common side]
ED = EB           [Tangent drawn from an external point to the line circle are equal to length]
$AE{O}'B\cong E{O}'D$   [By SSS congruence criterion]
$\Rightarrow \; \angle {O}'ED= \angle {O}'EB\, \cdots \left ( i \right )$
i.e.,      ${O}'E$ is bisector of $\angle BED$
Similarly, OE is the bisector of $\angle AEC$
In quadrilateral $DEB{O}'$
$\angle {O}'DE= \angle {O}'BE= 90^{\circ}$
$\Rightarrow \, \angle {O}'DE= \angle {O}'BE= 180^{\circ}$
$\therefore \angle DEB+\angle D {O}'B= 180^{\circ}\, \cdots \left ( ii \right )$ [ $\mathbb{Q}DEB{O}'$ is cyclic quadrilateral]
$\angle AED+\angle DEB= 180^{\circ}$ [ $\mathbb{Q}$ AB is a straight line]
$\angle AED+180^{\circ}-\angle D{O}'B= 180^{\circ}$ [From equation (ii)]
$\angle AED-\angle D{O}'B= 0$
$\angle AED-\angle D{O}'B\: \: \cdots \left ( iii \right )$
Similarly $\angle AED= \angle ADC\: \:\cdots \left ( iv \right )$
From equation (ii) $\angle DEB= 180^{\circ}-\angle D{O}'B$
Dividing both sides by 2
$\frac{\angle DEB}{2}= \frac{180^{\circ}-\angle D{O}'B}{2}$
$\angle DE{O}'= 90^{\circ}-\frac{1}{2}\angle D{O}'B\: \: \cdots \left ( v \right )$
$\left [ \because \, \frac{\angle DEB}{2}= \angle DE{O}' \right ]$
Similarly $\angle AEC= 180^{\circ}-\angle AOC$
Dividing both sides by 2
$\frac{1}{2}\angle AEC= 90^{\circ}-\frac{\angle AOC}{2}$
$\angle AEO= 90^{\circ}-\frac{1}{2}\angle AOC\; \cdots \left ( vi \right )$
$\left [ \because \frac{1}{2}\angle AEC= \angle AEO \right ]$
Now $\angle AEO+\angle AED+\angle DE{O}'$
$= 90-\frac{1}{2}\angle AOC+\angle AED+90^{\circ}-\frac{1}{2}\angle D{O}'B$
$= \angle AED+180^{\circ}-\frac{1}{2}\left ( \angle D{O}'B+\angle AOC \right )$
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ \angle AED+\angle AED \right ]$ [from equation (iii) and (iv)]
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ 2\angle AED \right ]$
$\angle AED+180^{\circ}-\angle AED$
$= 180^{\circ}$
$\therefore \: \angle AED+\angle DE{O}'+\angle AEO= 180^{\circ}$
So, OEO’ is a straight line
$\therefore$ O, E and ${O}'$ are collinear.
Hence Proved

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#9.4

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc.

Let the mid-point of the arc be C and DCE be the tangent to the circle.
Construction: Join AB, AC and BC.
Proof: In $\bigtriangleup$ ABC
AC = BC
$\Rightarrow \, \angle CAB= \angle CBA\: \cdots \left ( i \right )$
[ sides opposite to equal angles are equal]
Here DCF is a tangent line
$\therefore \; \; \angle ACD= \angle CBA$ [ angle in alternate segments are equal]
$\Rightarrow \; \angle ACD= \angle CAB$ …..(ii) [From equation (i)]
But Here $\angle ACD$ and $\angle CAB$ are alternate angels.
$\therefore$ equation (ii) holds only when AB||DCE.

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

Hence Proved.

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#9.4

AB is a diameter and AC is a chord of a circle with centre O such that $\angle$ BAC = $30^{\circ}$ . The tangent at C intersects extended AB at point D. Prove that BC = BD.

Given AB is the diameter and AC is a chord of a circle with centre O.
$\angle BAC= 30^{\circ}$
To Prove: BC = BD
Construction: Join B and C
Proof : $\angle BCD= \angle CAB$ [Angle is alternate segment]
$\angle BCD= 30^{\circ}\: \left [ Given \right ]$
$\angle BCD= 30^{\circ}\cdots \left ( i \right )$
$\angle ACB= 90^{\circ}$ [Angle in semi-circle formed is 90°]
In $\bigtriangleup$ ABC
$\angle CAB+\angle ABC+\angle BCA= 180^{\circ}$ [Sum of interior angles of a triangle is 180°]
$30^{\circ}+\angle ABC+90^{\circ}= 180^{\circ}$
$\angle ABC= 180^{\circ}-90^{\circ}-30^{\circ}$
$\angle ABC= 60^{\circ}$
Also, $\angle ABC+\angle CBD= 180^{\circ}$ [Linear pair]
$\angle CBD= 180^{\circ}-60^{\circ}$
$\angle CBD= 120^{\circ}$
$\angle ABC= 60^{\circ}$
In $\bigtriangleup CBD$
$\angle CBD+\angle BDC+\angle DCB= 180^{\circ}$
$\angle BDC= 30^{\circ}\cdots \left ( ii \right )$
From equation (i) and (ii)
$\angle BCD= \angle BDC$
$\Rightarrow BC= BD$ [ $\because$ Sides opposite to equal angles are equal]
Hence Proved

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#9.4

In Figure, tangents PQ and PR are drawn to a circle such that $\angle$ RPQ = $30^{\circ}$ . A chord RS is drawn parallel to the tangent PQ. Find the $\angle$ RQS.

In the given figure PQ and PR are two tangents drawn from an external point P.
PQ = PR [ $\mathbb{Q}$ lengths of tangents drawn from an external point to a circle are equal]
$\Rightarrow \angle PQR=\angle QRP$

[ angels opposite to equal sides are equal]
In $\bigtriangleup$ PQR
$\angle PQR+\angle QRP+\angle RPQ= 180^{\circ}$
[ sum of angels of a triangle is 180°]
$\angle PQR+\angle PQR+\angle RPQ= 180^{\circ}$
$\left [ \angle PQR= \angle QRP \right ]$
$2\angle PQR+30^{\circ}= 180^{\circ}$
$\angle PQR= \frac{180^{\circ}-30^{\circ}}{2}$
$\angle PQR= \frac{150^{\circ}}{2}$
$\angle PQR= 75^{\circ}$

SR||OP (Given)

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

Also $\angle PQR= \angle QRS= 75^{\circ}$ [Alternate segment angles]
In $\bigtriangleup$ QRS
$\angle Q+\angle R+\angle S= 180^{\circ}$
$\angle Q+75^{\circ}+75^{\circ}= 180^{\circ}$
$\angle Q= 180^{\circ}-75^{\circ}-75^{\circ}$
$\angle Q= 30^{\circ}$
$\therefore \angle RQS= 30^{\circ}$

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#9.4

In a right triangle ABC in which $\angle$ B = $90^{\circ}$ , a circle is drawn with AB as the diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

Let O be the centre of the given circle.
Suppose P meets BC at point R
Construction: Joinpoints P and B
To Prove: BR = RC
Proof: $\angle ABC= 90^{\circ}$ [Given]
In $\bigtriangleup$ ABC
$\angle ABC+\angle BCA+\angle CAB= 180^{\circ}$ [ Sum of interior angle of a triangle is 180°]
$90^{\circ}+\angle 2+\angle 1= 180^{\circ}$
$\angle 1+\angle 2= 180^{\circ}-90^{\circ}$
$\angle 1+\angle 2= 90^{\circ}$
Also     $\angle 4= \angle 1$          [ tangent and chord made equal angles in alternate segments]
$\Rightarrow \angle 4+\angle 2= 90^{\circ}\cdots \left ( i \right )$
$\angle APB= 90^{\circ}$    [ angle in semi-circle formed is 90°]
$\angle APB+\angle BPC= 180^{\circ}$
$\angle BPC= 180^{\circ}-90^{\circ}$
$\angle BPC= 90^{\circ}$
$\angle 4+\angle 5= 90^{\circ}\cdots \left ( ii \right )$
Equal equation (i) and (ii) we get
$\angle 4+\angle 2= \angle 4+\angle 5$
$\angle 2= \angle 5$
$PR= RC \, \cdots \left ( iii \right )$    [Side opposite to equal angles are equal]
Also,    PR = BR        …..(iv)   [ tangents drawn to a circle from external point are equal]

From equation (iii) and (iv)
BR = RC

Hence Proved

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#9.4

Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

Given: Radii of two circles are OP = 3 cm and ${O}'$ an intersection point of two circles are P and Q. Here two tangents drawn at point P are OP and ${O}'$ P.
$\therefore \: \angle P= 90^{\circ}$
$3^{2}-x^{2}= \left ( NP \right )^{2}$
Also, applying Pythagoras' theorem $\bigtriangleup PN{O}'$ we get
$\left ( P{O}' \right )^{2}= \left ( PN \right )^{2}+\left (N {O}' \right )^{2}$
$\left (4 \right )^{2}= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16- \left (5-x \right )^{2}= \left ( PN \right )^{2}\cdots \left ( ii \right )$
Equate equation (i) and (ii) we get
$9-x^{2}= 16-\left ( 5-x \right )^{2}$
$9-x^{2}-16+\left ( 25+x^{2}-10x \right )= 0$
$\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]$
$-7-x^{2}+\left ( 25+x^{2}-10x \right )= 0$
$-7-x^{2}+25+x^{2}-10x= 0$
$18-10x= 0$
$18= 10x$
$\frac{18}{10}= x$
$x= 1\cdot 8$
Put x = 1.8 in equation (i) we get
$9-\left ( 1\cdot 8 \right )^{2}= NP^{2}$
$9-3\cdot 24= NP^{2}$
$5\cdot 76= NP^{2}$
$NP= \sqrt{5\cdot 76}$
$NP= 2\cdot 4$
$P\mathbb{Q}= 2\times PN= 2\cdot 24= 4\cdot 8 cm$