Question

Two circles having a common tangent at a and b . If OA = 12 CM and O'b = 3 cm then find AB

Answer 1

∠AEC = ∠DEB (Vertically Opposite Angle)
join OA and OC
so in triangle OAE and triangle OCE we have,
OA=OC (radii of same circle)
OE=OE (common)
∠OAE=∠OCE (90° as the tangent is always perpendicular to the radius at the                            point of contact)
∴ ΔOAE ≡ ΔOCE
so ∠AEO = ∠CEO (CPCT)
similarly for other circle we have
∠DEO' = ∠BEO' (CPCT)
Now ∠AEC=∠DEB
⇒1/2(∠AEC) = 1/2(∠DEB)
⇒∠AEO =∠CEO = ∠DEO'= ∠BEO'
so all 4 ∠'s are equal and bisected by OE and OE'
∴ O,E,O' are collinear 

.......................OR............................... OR.........................,

Given : AB and CD are common tangents to circles with centre O and O'  intersecting at E

We know that tangents drawn from an exterior point are equally inclined to the segment joining the centre to that point

⇒ ∠AEO = ∠CEO

or

EO is the angle bisector of ∠AEC  ...  (1)

 

and ∠BEO' = ∠DEO'

or

EO' is the angle bisector of ∠BED  ...  (2)

 

but ∠AEC = ∠BED  ...  (3)

 

from (1), (2),(3),

E, O', O are collinear.