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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\[\frac{DO'}{CO}\]

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#### Solution

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`

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