1 of 3

CIRCLE

Sum based on Theorems –
Two tangents from an external point

to a circle are equal and

Radius is perpendicular to the tangent

2 of 3

[Length of the tangents

drawn from an external

point to a circle are equal]

T

O

P

Q

Proof :

In ΔPTQ,

= 180^o

∠TPQ

∠TQP

=

PT

=

TQ

∠TPQ

∠TQP

+

∠PTQ

+

= 180^o

∠TPQ

+

∠PTQ

+

= 180^o

2∠TPQ

∠PTQ

+

= 180^o

2∠TPQ

∠PTQ

–

T

We know, sum of all angles of a triangle is 180⁰

Two tangents TP and TQ are drawn to a circle with centre O

from an external point T.

Prove that ∠PTQ = 2 ∠OPQ

Consider ΔPTQ

[Angle sum property]

…(i)

[from (i)]

…(ii)

∴

[Angles opposite to equal sides]

∴

3 of 3

[From (ii) and (iii)]

= 90^o

∠TPQ

∠OPQ

+

Proof :

∠OPT

90^o

=

= ∠OPT

∠TPQ

∠OPQ

+

= 90^o

∠TPQ

∠OPQ

–

∠PTQ

=

2 ∠OPQ

2∠OPQ

∠OPT is made up of two angles

i.e. ∠OPQ and

∠TPQ

= 180^o

2∠TPQ …(ii)

∠PTQ

–

= 180^o

2∠TPQ

–

T

O

P

Q

Two tangents TP and TQ are drawn to a circle with centre O from an external point T.

Prove that ∠PTQ = 2 ∠OPQ

[Radius is perpendicular to

tangent]

…(iii)

∴

We know, radius is perpendicular to tangent

∠OPQ is a part of ∠OPT

Let us multiply throughout by 2

∠OPT = 90⁰

∴