Question

In the below figure, two equal circles, with Centre’s \[{\text{O}}\] and \[{{\text{O}}'}\], touch each other at \[{\text{X}}\]. \[{{\text{O}}'}{\text{X}}\] produced meets the circle with center \[{\text{O}}\] at \[{\text{A}}\]. \[{\text{AC}}\] is tangent to the circle with Centre \[{\text{O}}\] at the point \[{\text{C}}\]. \[{{\text{O}}^{\text{'}}}{\text{D}}\] is perpendicular to \[{\text{AC}}\]. Find the value of \[\dfrac{{{\text{D}}{{\text{O}}^{\text{'}}}}}{{{\text{CO}}}}\].

Answer 1

Hint: Here, we will use the property of tangents which states that any tangents drawn to the circle at any point are perpendicular to the radius of the circle through the point of contact.
We will also use the similarity property here which defines that, if the two sides of a triangle are in the same proportion of the two sides of another triangle, and the angle inscribed by the two sides in both the triangles are equal, then two triangles are said to be similar. The symbol of similarity is ~.

Complete step-by-step solution:
Step 1: As shown in the above figure (1), \[{\text{AC}}\] is tangent to circle having a center at \[{\text{O}}\], now as we know that tangents that are drawn to the circle make an angle of \[{\text{9}}{{\text{0}}^0}\] with the radius of the circle, so we can say that:
 \[ \Rightarrow \angle {\text{ACO = 9}}{{\text{0}}^0}\]………… (1)
It is given in the question that \[{{\text{O}}^{\text{'}}}{\text{D}}\] is perpendicular to \[{\text{AC}}\], so we have:
\[ \Rightarrow \angle {\text{ADO' = 9}}{{\text{0}}^0}\]………….(2)
Now, from the expressions (1) and (2), both the angles are equal to \[9{{\text{0}}^0}\] and so \[\angle {\text{ADO'}} = \angle {\text{ACO}} = 9{{\text{0}}^0}\].
Therefore, by using the principle of the corresponding angle which states that the pair of the angles which is on the same side of one of two lines are equal then, the two lines which are intersected by a transversal are parallel. So,
\[ \Rightarrow {\text{DO'}}\parallel {\text{CO}}\]
Now, \[{\text{DO'}}\parallel {\text{CO}}\] so, by using the principle of alternate angles, which states that if the two lines are parallel and intersect by a transversal than the alternate interior angles are equal. So:
\[ \Rightarrow \angle {\text{AO'D = }}\angle {\text{AOC}}\].
Step 2: Now, in \[\Delta {\text{ADO'}}\] and \[\Delta {\text{ACO}}\]:
\[\angle {\text{ADO'}} = \angle {\text{ACO}} = 9{{\text{0}}^0}\] (by equation (1) and (2))
 Also, from the figure (1):
\[\angle {\text{DAO'}} = \angle {\text{CAO}}\] \[\because \] (the angle is common)
\[\angle {\text{AO'D = }}\angle {\text{AOC}}\] \[\because \] (already proved in step 1)
Therefore, by using the AAA (Angle-Angle-Angle) property of similarity ie all three angles are equal both the triangles are similar:
 \[ \Rightarrow \Delta {\text{ADO'}} ~ \Delta {\text{ACO}}\]
Step 3: By using the information given in the question, the two circles are equal so the radii of both of the circles are the same and therefore \[{\text{AO'}} = {\text{O'X}} = {\text{XO}}\].
Now from the above figure (1), we can see that \[{\text{AO = AO' + O'X + XO}}\].
Thus, for finding the ratio of \[\dfrac{{{\text{AO'}}}}{{{\text{AO}}}}\] by substituting \[{\text{AO = AO' + O'X + XO}}\] in \[\dfrac{{{\text{AO'}}}}{{{\text{AO}}}}\]:
\[ \Rightarrow \dfrac{{{\text{AO'}}}}{{{\text{AO}}}} = \dfrac{{{\text{AO'}}}}{{{\text{AO'}} + {\text{O'X}} + {\text{XO}}}}\]
Now, by using \[{\text{AO'}} = {\text{O'X}} = {\text{XO}}\] in \[\dfrac{{{\text{AO'}}}}{{{\text{AO}}}} = \dfrac{{{\text{AO'}}}}{{{\text{AO'}} + {\text{O'X}} + {\text{XO}}}}\] we get:
\[ \Rightarrow \dfrac{{{\text{AO'}}}}{{{\text{AO}}}} = \dfrac{{{\text{AO'}}}}{{{\text{AO'}} + {\text{AO' + AO'}}}}\]
By adding the denominator in the RHS side and dividing it by the numerator we get:
\[ \Rightarrow \dfrac{{{\text{AO'}}}}{{{\text{AO}}}} = \dfrac{{\text{1}}}{3}\]……….(3)
Step 4: From step number 2, \[\Delta {\text{ADO'}} ~ \Delta {\text{ACO}}\] so by using the property of similarity which states that if two triangles are similar then their ratio of corresponding sides are also equal we have:
 \[\dfrac{{{\text{DO'}}}}{{{\text{CO}}}} = \dfrac{{{\text{AO'}}}}{{{\text{AO}}}}\]
Therefore, by using the result \[\dfrac{{{\text{AO'}}}}{{{\text{AO}}}} = \dfrac{1}{3}\]from equation (3), and \[\dfrac{{{\text{DO'}}}}{{{\text{CO}}}} = \dfrac{{{\text{AO'}}}}{{{\text{AO}}}}\], we get:
\[ \Rightarrow \dfrac{{{\text{DO'}}}}{{{\text{CO}}}} = \dfrac{1}{3}\].

So \[\dfrac{{{\text{DO'}}}}{{{\text{CO}}}}\] is equal to \[\dfrac{1}{3}\]

Note: In these types of questions, students often get confused while applying the similarity properties of triangles so, you should remember all the rules of similar triangles and apply them correctly which we have used multiple times above as:
In step number 2: By using the AAA (Angle-Angle-Angle) property of similarity both the triangles are similar.
In step number 4: By using the property of similarity which states that if two triangles are similar then their ratio of corresponding sides is also equal.
You should also remember the properties of corresponding angles and alternate angles which we have used in step number 1.