Question

The common chord of two intersecting circles \[{c_1}\] & \[{c_2}\] can be seen from their centers at the angles of \[90^\circ \] and \[60^\circ \] respectively. If the distance between their centers is equal to \[\sqrt 3 + 1\], then the radii of \[{c_1}\] & \[{c_2}\] are
A. \[\sqrt 3 \] & 3
B. \[\sqrt 2 \] & \[2\sqrt 2 \]
C.\[\sqrt 2 \] & 2
D. \[2\sqrt 2 \] & 4

Answer 1

Hint:
Here, first we will draw the diagram for the question. To make the diagram we have to use the information given in the question and label the vertices. Then solve for both the triangle separately and use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the triangle.

Complete step by step solution:
We are given that the distance between circles \[{O_1}{O_2}\] is \[\sqrt 3 + 1\].
We will take the length \[A{O_1}\] is \[{c_1}\] and \[A{O_2}\] is \[{c_2}\].
Let us assume that the distance \[{O_1}C\] is \[x\].
Then, the length \[{O_2}C\] will be \[\sqrt 3 + 1 - x\].

First, we will take the triangle \[\Delta AC{O_1}\].
\[
  \angle A{O_1}C = \dfrac{1}{2}\angle A{O_1}B \\
   = 45^\circ \\
 \]
We will find the value of \[AC\] using the tangential function \[\tan 45^\circ \].
\[
   \Rightarrow \tan 45^\circ = \dfrac{{AC}}{{{O_1}C}} \\
   \Rightarrow 1 = \dfrac{{AC}}{x} \\
   \Rightarrow AC = x{\text{ ......}}\left( 1 \right) \\
 \]
Using the Pythagorean theorem in \[\Delta AC{O_1}\], we get
\[
  A{O_1}^2 = A{C^2} + C{O_1}^2 \\
  {c_1}^2 = {x^2} + {x^2} \\
  {c_1}^2 = 2{x^2} \\
  {c_1} = x\sqrt 2 \\
 \]
Now, we will take the triangle \[\Delta AC{O_2}\].
\[
  \angle A{O_2}C = \dfrac{1}{2}\angle A{O_2}B \\
   = \dfrac{1}{2} \times 60^\circ \\
   = 30^\circ \\
 \]
We will find the value of \[AC\] using the tangential function \[\tan 30^\circ \].
\[
   \Rightarrow \tan 30^\circ = \dfrac{{AC}}{{{O_2}C}} \\
   \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{AC}}{{\sqrt 3 + 1 - x}} \\
   \Rightarrow AC = \dfrac{{\sqrt 3 + 1 - x}}{{\sqrt 3 }}{\text{ ......}}\left( 2 \right) \\
 \]
From \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[
   \Rightarrow x = \dfrac{{\sqrt 3 + 1 - x}}{{\sqrt 3 }} \\
   \Rightarrow \sqrt 3 x = \sqrt 3 + 1 - x \\
   \Rightarrow \sqrt 3 x + x = \sqrt 3 + 1 \\
   \Rightarrow \left( {\sqrt 3 + 1} \right)x = \sqrt 3 + 1 \\
   \Rightarrow x = 1 \\
 \]
Substitute this value of \[x\] in \[AC\], \[A{O_1}\] and \[{O_2}C\].
\[ \Rightarrow AC = 1\]
\[
   \Rightarrow A{O_1} = x\sqrt 2 \\
   \Rightarrow {c_1} = \sqrt 2 \\
   \Rightarrow {O_2}C = \sqrt 3 + 1 - 1 \\
   \Rightarrow {O_2}C = \sqrt 3 \\
 \]
Using these values in the Pythagorean theorem in \[\Delta AC{O_2}\], we get
\[
  A{O_2}^2 = A{C^2} + {O_2}{C^2} \\
  {c_2}^2 = {1^2} + {\left( {\sqrt 3 } \right)^2} \\
  {c_2}^2 = 1 + 3 \\
  {c_2}^2 = 4 \\
  {c_2} = 2 \\
 \]

Thus, the value of radii \[{c_1}\] is \[\sqrt 2 \] and \[{c_2}\] is \[2\].
Hence, option C is correct.

Note:
In this question, the most important thing is to draw the diagram using the information given in the question to avoid miscalculation and label the points carefully. You should be familiar with the basic properties of the circle. Also, we will write the values of the sides properly in the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the triangle.