Question

Prove that the angle between the two tangents drawn from an external point to a circle
is supplementary to the angle subtended by
of contact at the Joining the points centre, to the
the segment . ​

Answer 1

Complete Question :

›»› Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Given :

• PA and PB are the tangents to the circle whose centre is O.
• A and B are the points of contact.

To Prove :

• The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Construction :

• Join OA and OB.

Proof :

Here, PA and PB are the tangents to the circle whose centre is O.

A and B are the points of contact.

Since PA is a tangent and AO is a radius,

→ ∠1 = 90° ...........(1)

Similarly, PB is a tangent and BO is a radius,

→ ∠2 = 90° .............(2)

Our equation are,

• ∠1 = 90° ...........(1)
• ∠2 = 90° .............(2)

Adding equation (1) and equation (2), we get:

→ ∠1 + ∠2 = 90° + 90°

→ ∠1 + ∠2 = 180°.

Now, in quadrilateral AOBP, we have:

→ ∠1 + ∠P + ∠2 + ∠O = 360°

→ ∠1 + ∠2 + ∠P + ∠O = 360°

→ 180° + ∠P + ∠O = 360°

→ ∠P + ∠O = 360° - 180°

→ ∠P + ∠O = 180°

i,e, ∠P and ∠O are supplementary angle.

Hence Proved !

Answer 2

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Given :

• the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by of contact at the Joining the points centre

To Prove :

• Prove that the angle between the two tangents drawn from an external point

Solution :

Let us consider a circle centered at point O. P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 90°

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 90°

We know that :

Sum of all interior angles = 360º

Substitute all values :

∠OAP +∠APB+∠PBO +∠BOA = 360º

90º + ∠APB + 90º + ∠BOA = 360º

∠APB + ∠BOA = 360° - 180°

∠APB + ∠BOA = 180º

Hence, proved !!