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## All Questions

#### 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

#### 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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#### 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

#### 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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#### 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.

#### 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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#### 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.

#### 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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#### 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

#### 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.

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.