[1]If PA and PB are tangents from an outside point P such that PA = 10 cm and angle APB = 60 .find the length of the chord AB.

[2]prove that the tangent drawn at the ends of a chord of a circle make equal with the chord.

[3]prove that the line segment joining the points of contact of two parallel tangent to a circle is a diameter of the circle.

Dear Student!

Here is the answer to your query.

 

(1) Given : PA and PB are tangents of a circle, PA = 10 cm and ∠APB = 60°

Let O be the center of the given circle and C be the point of intersection of OP and AB

In ΔPAC and ΔPBC

PA = PB  ( Tangents from an external point are equal)

∠APC = ∠BPC ( Tangents from an external point are equally inclined to the segment joining center to that point)

PC = PC ( Common)

Thus ΔPAC ΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

∠ACP + ∠BCP = 180°

Now in right triangle ACP

∴ AB = AC + BC = AC + AC  ( AC = BC)

⇒ AB = (5 + 5) cm = 10 cm

 

(2) From (1) we have 

∠PAC = ∠PBC

 

(3)

 

Given : Tangent l is parallel to tangent m end A and B are its points of intersection with circle of center O and radius r.

To Prove : AB is the diameter

Let l and m meet at some point P

Then in quadrilateral AOBP

∠AOB + ∠OAP + ∠OBP + ∠APB = 360° ......(2)

We know that tangents to a circle is perpendicular to the radius

⇒ OA ⊥ l and OB ⊥ m

⇒ ∠OAP = 90° and ∠OBP = 90° .........(3)

Since l ||

∠APB = 0° ........(4)

From (2), (3) and (4)

∠AOP + 90° + 90° + 0° = 360°

⇒ ∠AOB = 360° – 180° = 180°

AB = AO + OB = r + r = 2r = Diameter 

Hence AB is the diameter of the circle.

 

Cheers!

  • 49

Thankx sir for ur help......

  • -1

Sir may i ask u a dougbt ??

Where is P in this diagram ? plz explain 

  • 8
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