Equal circles with centres O and O' touch each other at X. OO' produced to meet


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 $\frac{D O^{\prime}}{C O}$.


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)


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 OX are the radii of the same circle, we have,


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


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,

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


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

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