**Answer
1** :

PT is the tangent to the circle with centre O, at T

Radius OT = 8 cm, OP = 17 cm

PT is the tangent segment

Now in right ∆OPT,

OP² = OT² + PT² (Pythagoras Theorem)

=> (17)² = (8)² + PT²

=> 289 = 64 + PT²

=> PT² = 289 – 64 = 225 = (15)²

PT = 15 cm

**Answer
2** :

From a point P outside the circle with centre O, PT is the tangent to the circle and radius

OT = 5 cm, OP = 15 cm

OT ⊥ PT

Now in right ∆OPT,

OP² = OT² + PT² (Pythagoras Theorem)

(13)² = (5)² + PT²

=> 169 = 25 + PT²

=> PT² = 169 – 25 = 144 = (12)²

PT = 12 cm

**Answer
3** :

From a point P outside the circle of centre 0 and radius OT, PT is the tangent to the circle

OP = 26 cm, PT = 10 cm

Now in right ∆OPT

Let r be the radius

OP² = OT² + PT² (Pythagoras Theorem)

=> (26)² = r² + (10)²

=> 676 = r² + 100

=> 676 – 100 = r²

=> r² = 576 = (24)²

r = 24

Hence radius of the circle = 24 cm

**Answer
4** :

Given : QR is the common chord of two circles intersecting each other at Q and R

P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centres O and C respectively

To prove : PT = PS

Proof: PT is the tangent and PQR is the secant to the circle with centre O

PT² = PQ x PR ….(i)

Similarly PS is the tangent and PQR is the secant to the circle with centre C

PS² = PQ x PR ….(ii)

From (i) and (ii)

PT² = PS²

PT = PS

Hence proved.

**Answer
5** :

Given : The sides of a quadrilateral ABCD touch the circle at P, Q, R and S respectively

To prove : AB + CD = AP + BC

Proof : AP and AS are the tangents to the circle from A

AP = AS ….(i)

Similarly BP = BQ ……(ii)

CR = CQ ….(iii)

and DR = DS ….(iv)

Adding, we get

AP + BP + CR + DR = AS + BQ + CQ + DS

=> (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

Hence proved.

**Answer
6** :

Let C1 and C2 be the two circles having same centre O. AC is a chord which touches the C1 at point D

Join OD.

Also, OD ⊥ AC

AD = DC = 4 cm

[perpendicular line OD bisects the chord]

In right angled ∆AOD,

OA² = AD² + DO²

[by Pythagoras theorem, i.e.,

(hypotenuse)² = (base)² + (perpendicular)²]

=> DO² = 5² – 4² = 25 – 16 = 9

=> DO = 3 cm

Radius of the inner circle OD = 3 cm

**Answer
7** :

Given : Chord PQ is parallel tangent at R.

To prove : R bisects the arc PRQ.

Proof: ∠1 = ∠2 [alternate interior angles]

∠1 = ∠3

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

∠2 = ∠3

=> PR = QR

[sides opposite to equal angles are equal]

=> PR = QR

So, R bisects PQ.

**Answer
8** :

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.

∠MAO = 90°

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

∠CEO = ∠MAO [corresponding angles]

∠CEO = 90°

Thus, OE bisects CD

[perpendicular from centre of circle to chord bisects the chord]

Similarly, the diameter AB bisects all. Chord which are parallel to the tangent at the point A.

**Answer
9** :

Given : AB, AC and PQ are the tangents to the circle as shown in the figure above and AB = 5 cm

To find : The perimeter of ∆APQ

Proof: PB and PX are the tangents to the circle

PB = PX

Similarly QC and QX are the tangents from

QC = QX

and AB and AC are the tangents from A

AB = AC

Now perimeter of ∆APQ

= AP + PQ + AQ

= AP + PX + QX + AQ

= AP + PB + QC + AQ { PB = PX and QC = QX}

= AB + AC

= AB + AB (AB=AC)

= 2 AB = 2 x 5 = 10 cm

**Answer
10** :

Given : PQ and RS are parallel tangents of a circle

RMP is the intercept of the tangent between PQ and RS

RO and PQ are joined

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