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.
prove that the tangent drawn at the ends of a chord of a circle make equal with the chord.
prove that the line segment joining the points of contact of two parallel tangent to a circle is a diameter of the circle.
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
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 || m
∴ ∠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.