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#9.3

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

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 parallel to the tangent at 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.

According to question

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

Take a point C and join AC and AB.
Now $\angle Q_{1}= \angle C$ …..(i) (Angle in the 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.

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 the angle between the tangent and chord is equal to the angle made by the chord in the 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 are 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.

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

To Prove AB = CD
According to question

It is given that the radius of both circles is 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 sides 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.

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

Here EA = EC ( $\because$ length of tangent drawn from the 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.

Here PQ and PR are tangents and O is the center of the 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 lies on the 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 and BD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.

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 bisects $\angle DBC$
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.

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 the sum of the interior angles of the 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 the sum of opposite angles of the 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.

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 the 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 parallel to the tangent at 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 the above question, if the 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 the above question, if the 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 and BD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.