**Question:**

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

**Solution:**

Consider the two triangles and .

We have,

is a common angle for both the triangles.

(Given in the problem)

(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,

From *AA* similarity postulate we can say that,

~

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 and *AO* of .

$\frac{A O^{\prime}}{A O}=\frac{A O^{\prime}}{A O^{\prime}+O^{\prime} X+O X}$

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

$\frac{A O^{\prime}}{A O}=\frac{A O^{\prime}}{A O^{\prime}+A O^{\prime}+O^{\prime} A}$

$\frac{A O^{\prime}}{A O}=\frac{1}{3}$

Since $\triangle A C O \sim \Delta A D O^{\prime}$,

$\frac{A O^{\prime}}{A O}=\frac{D O}{C O}$

We have found that,

$\frac{A O^{\prime}}{A O}=\frac{1}{3}$

Therefore,

$\frac{D O^{\prime}}{C O}=\frac{1}{3}$

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