CIRCLE

• Theorem – The lengths of two tangents

drawn from an external point to a circle are equal.

THEOREM :

The lengths of two tangents drawn from an external

point to a circle are equal.

O

Given : (i) A circle with centre O.

(ii) P is a point in the exterior of the circle.

(iii) PA and PB are the tangents from P.

P

A

B

To prove :

PA = PB

Construction :

Draw OA, OB

Proof :

In ΔPAO and ΔPBO,

∠PAO = ∠PBO = 90^o

[Radius is perpendicular to the tangent]

Hypt. OP = Hypt. OP

[Common side]

II

I

I

OA = OB

[radii of the same circle]

∴ ΔPAO ≅ ΔPBO

[RHS rule]

∴ PA = PB

[c.p.c.t]

Now, do we have one more pair of sides equal ?

Yes

OA = OB

Now, let us check whether the hypotenuse of triangles are equal?

Yes

OP = OP

Whenever we see a centre

and point of contact…

Draw radius

For proving sides equal,

prove triangles congruent

Do we see triangles ?

No

Let us create triangles

by drawing OP

and OP

A

B

C

P

Q

R

AP = AR

BP = BQ

CQ = CR

[The lengths of two tangents drawn

from an external point to a circle are equal]

How many external points do you observe ?

3

A, B and C

Let us consider point A

Name the tangents from external point A

AP and AR

Let us consider point B

Name the tangents from external point B

BP and BQ

Let us consider point C

Name the tangents from external point C

CQ and CR

What will be the reason ?

PA = PD

QA = QB

RB = RC

SC = SD

How many external points do you observe ?

4

P, Q, R and S

Let us consider point P

Name the tangents from external point P

PA and PD

Let us consider point Q

Name the tangents from external point Q

QA and QB

Let us consider point R

Name the tangents from external point R

RB and RC

What will be the reason ?

Let us consider point S

Name the tangents from external point S

SC and SD

[The lengths of two tangents drawn from

an external point to a circle are equal]