Question

Equal circles with centres O and O' touch each other at X. OO' produced to meet a circle with centre O', at A. AC is a tangent to the circle whose centre is O. O'D is perpendicular to AC. Find the value of\[\frac{DO'}{CO}\]

Answer 1

Consider the two triangles Δ ADOand Δ ACO.

We have,

∠ A is a common angle for both the triangles.

`∠ ADO =90^o` (Given in the problem)

`∠ ACO = 90^o` (Since OC is the radius and AC is the tangent to that circle at C and we know that the radius is always perpendicular to the tangent at the point of contact)

Therefore,

`∠ ADO = ∠ACO`

From AA similarity postulate we can say that,

Δ ACO ~ Δ ADO

Since the triangles are similar, all sides of one triangle will be in same proportion to the corresponding sides of the other triangle.

Consider AO′ of Δ ADO  and AO of  Δ ACO .

`(AO)/(AO)=(AO)/(AO+OX+OX)`

Since AO′ and O′X are the radii of the same circle, we have,

AO′ = O′X

Also, since the two circles are equal, the radii of the two circles will be equal. Therefore,

AO′ = XO

Therefore we have

`(AO)/(AO)=(AO)/(AO+AO+AO)`

`(AO)/(AO)=1/3`

Since Δ ACO ~ Δ ADO,

`(AO)/(AO)=(DO)/(CO)`

We have found that,

`(AO)/(AO)=1/3`

Therefore,

`(DO)/(CO)=1/3`