Question

if from any point P on the common tangent of two circles touching each other externally,tangents PA and PB are drawn to the circles.If PA=5cm, find the length of the tangent PB

Answer 1

Solution:

Theorem that is used in this question is : length of tangents from external point to a circle are equal.

It is given that , from point P  on the common tangent of two circles touching each other externally,tangents PA and PB are drawn to the circles.

Also, PA= 5 cm

PA= PM= 5 cm→→[ length of tangents from external point to a circle are equal.]------(1)

PM= PB= 5 cm→→[ length of tangents from external point to a circle are equal.]--------(2)

Equating (1) and (2)

So, Length of tangent PB= Length of tangent PA=5 cm

Answer 2

From figure, PA and PB are the tangents.

O is the centre of the circle.

To Prove : AOBP is a cyclic quadrilateral

Now,

OA is radius and PA is tangent

OA⊥PA

So, ∠OAP=90

∘

___(1)

Similarly, OB is radius and PB is tangent.

OB⊥PB

So, ∠OBP=90

∘

__(2)

Add (1) and (2), we have

∠OAP+∠OBP=90

∘

+90

∘

=180

∘

But these are opposite angles of the quadrilateral AOBP.

Therefore, Quadrilateral AOBP is a cyclic....

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