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#9.3

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Solution
According to question

Let us take a chord EF || XY
Here $\angle XAO= 90^{\circ}$
( $\because$ tangent at any point of the circle is perpendicular to the radius through the point of contact)

$\angle EGO= \angle XAO$ (Corresponding angles)

$\therefore \, \angle EGO= 90^{\circ}$
Thus AB bisects EF
Hence AB bisects all the chords which are parallel to the tangent at the point A.
Hence Proved

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#9.3

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Solution
According to question

To Prove : $\angle Q_{1}= \angle Q_{2}$
X₁, Y₁, X₂, Y₂ are tangents at point A and B respectively.

Take a point C and join AC and AB.
Now $\angle Q_{1}= \angle C$ …..(i) (Angle in alternate segment)
Similarly,
$\angle Q_{2}= \angle C$ …..(ii)
From equation (i) and (ii)
$\angle Q_{1}= \angle Q_{2}= \angle C$ …..(iii)
From equation (3)
$\angle Q_{1}= \angle Q_{2}$
Hence Proved

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#9.3

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Solution
According to question

To Prove : R bisects the arc PRQ
Here $\angle Q_{3}= \angle Q_{1}$ …..(i) ( $\because$ Alternate interior angles)
We know that angle between tangent and chord is equal to angle made by chord in alternate segment.

$\therefore \angle Q_{3}= \angle Q_{2}$ …..(ii)
From equation (i) and (ii)
$\angle Q_{1}= \angle Q_{2}$
We know that sides opposite to equal angles and equal.
$\therefore$ PR = QR
Hence R bisects the arc PRQ
Hence Proved

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#9.3

In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Solution
To Prove : AB = CD
We know that the length of tangents drawn from some point to a circle is equal.
So, EB = ED
AE = CE
Here AB = AE + EB
CD = CE + ED
From Euclid axiom is equals are added in equals then the result is also equal.
$\therefore$ AB = CD
Hence Proved

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#9.3

In Figure, AB and CD are common tangents to two circles of unequal radii.

In above question, if radii of the two circles are equal, prove that AB = CD.

Solution
To Prove AB = CD
According to question

It is given that radius of both circles are equal
Hence, OA = OC = PB = PD
Here, $\angle A= \angle B= \angle C= \angle D= 90^{\circ}$
( $\because$ tangent at any point is perpendicular to the radius at the point of contact)
Hence ABCD is a rectangle.
Opposite side of a rectangle are equal

$\therefore AB= CD$
Hence Proved.

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#9.3

In Figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.

Solution
To Prove : AB = CD
Extend AB and CD then they meet at point E.

Here EA = EC ( $\because$ length of tangent drawn from same point is equal)
Also EB = ED ( $\because$ length of tangent drawn from some point is equal)
AB = AE - BE
CD = CE -DF
From equal axiom if equal are subtracted from equal then the result is equal.

$\therefore AB= CD$

Hence Proved

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#9.3

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Solution

Here PQ and PR are tangents and O is the center of circle.
   Let us join OQ and OR.
Here $\angle OQP= \angle ORP= 90^{\circ}$
( $\because$ tangent from exterior point is perpendicular to the radius through the point of contact)
In $\bigtriangleup$ PQO and $\bigtriangleup$ PRO
$OQ= OR$ (Radius of circle)
$OP= OP$         (Common side)
Hence, $\bigtriangleup PQO\cong \bigtriangleup PRO$      [RHS interior]
Hence, $\angle RPO= \angle QPO$       [By CPCT]
Hence O lie on angle bisector of $\angle QPR$

Hence Proved.

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#9.3

If from an external point B of a circle with centre O, two tangents BC andBD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.

Solution
According to question

Given : $\angle DBC= 120^{\circ}$
To Prove : BC + BD = BO, i.e., BO = 2BC
Here OC $\perp$ BC, OD $\perp$ BD ( $\because$ BC, BD are tangents)
Join OB which bisect $\angle DBC$
In $\bigtriangleup ODB$
$\cos \theta = \frac{B}{H}$
$\cos 60^{\circ}= \frac{BD}{OB}$
$\frac{1}{2}= \frac{BD}{OB}$
$OB= 2BD$
$OB= 2BC$ $\left ( \because BD= BC \, \ \ Tangents \right )$
$OB= BC+BC$
$OB= BC+BD\; \; \left ( \because BC= BD \right )$
Hence Proved

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#9.3

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Solution
According to question

To Prove : QORP is a cyclic quadrilateral.
OQ $\perp$ PQ, OR $\perp$ PR ( $\because$ PQ, PR are tangents)
Hence, $\angle OQP+\angle ORP= 180^{\circ}\cdots (i)$
We know that sum of interior angles of quadrilateral is $360^{\circ}$
$\angle OPQ+\angle OPR+\angle ORP+\angle ROQ= 360^{\circ}$ [Given (i)]
$180^{\circ}+\angle QPR+\angle ROQ= 360^{\circ}$

$\angle QPR+\angle ROQ= 180^{\circ}$
Here we found that sum of opposite angles of quadrilateral is $180^{\circ}$
Hence QORP is a cyclic quadrilateral.
Hence proved

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#9.3

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer Radius = 3 cm
Solution
According to question

Given : AC = 8 CM
OC = 5 CM
AC = AB+BC
8 = BC+BC ( $\because$ AB =BC)
BC = 4CM
In $\bigtriangleup OBC$ using Pythagoras theorem
$H^{2}+B^{2}+P^{2}$
$\left ( OC \right )^{2}= \left ( BC \right )^{2}+\left ( OB \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( OB \right )^{2}$
$\left ( OB \right )^{2}= 25-16$
$OB= \sqrt{9}= 3cm$
Hence radius of inner circle is 3 cm.

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Get Answers to all your Questions

All Questions

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Posted by

infoexpert27

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Posted by

infoexpert27

In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Posted by

infoexpert27

JEE Main high-scoring chapters and topics

In Figure, AB and CD are common tangents to two circles of unequal radii. In above question, if radii of the two circles are equal, prove that AB = CD.

In Figure, AB and CD are common tangents to two circles of unequal radii.

In above question, if radii of the two circles are equal, prove that AB = CD.

If from an external point B of a circle with centre O, two tangents BC andBD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.