Let two circles having radii `r_(1) and r_(2)` are orthogonal to each other. If the length of their common chord is k times the square root of harmonic mean between the squares of their radii, then `k^(4)` is equal to

Options

A. 13
B. 7
C. 4
D. 2

Answer

To solve the problem step by step, let's break it down as follows:

### Step 1: Understanding the Problem
We have two circles with radii \( r_1 \) and \( r_2 \) that are orthogonal to each other. The length of their common chord is given to be \( k \) times the square root of the harmonic mean of the squares of their radii.

### Step 2: Finding the Length of the Common Chord
For two orthogonal circles, the length of the common chord \( AB \) can be calculated using the formula:
\[ AB = 2 \cdot \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}} \]

### Step 3: Finding the Harmonic Mean
The harmonic mean \( H \) of the squares of the radii \( r_1^2 \) and \( r_2^2 \) is given by:
\[ H = \frac{2 \cdot r_1^2 \cdot r_2^2}{r_1^2 + r_2^2} \]

### Step 4: Relating the Length of the Common Chord to \( k \)
According to the problem, we have:
\[ AB = k \cdot \sqrt{H} \]
Substituting the expressions we found:
\[ 2 \cdot \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}} = k \cdot \sqrt{\frac{2 \cdot r_1^2 \cdot r_2^2}{r_1^2 + r_2^2}} \]

### Step 5: Simplifying the Equation
Squaring both sides to eliminate the square roots gives:
\[ \left(2 \cdot \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}}\right)^2 = k^2 \cdot \left(\frac{2 \cdot r_1^2 \cdot r_2^2}{r_1^2 + r_2^2}\right) \]
This simplifies to:
\[ \frac{4 r_1^2 r_2^2}{r_1^2 + r_2^2} = k^2 \cdot \frac{2 r_1^2 r_2^2}{r_1^2 + r_2^2} \]

### Step 6: Canceling Common Terms
Assuming \( r_1^2 + r_2^2 \neq 0 \) and \( r_1^2 r_2^2 \neq 0 \), we can cancel \( r_1^2 r_2^2 \) from both sides:
\[ 4 = 2k^2 \]

### Step 7: Solving for \( k^2 \)
Dividing both sides by 2 gives:
\[ k^2 = 2 \]

### Step 8: Finding \( k^4 \)
To find \( k^4 \), we square \( k^2 \):
\[ k^4 = (k^2)^2 = 2^2 = 4 \]

### Final Answer
Thus, the value of \( k^4 \) is:
\[ \boxed{4} \]