If `(a ^(2) + b ^(2))/(c ^(2)) = (b ^(2) + c ^(2))/( a ^(2)) = (c ^(2) + a ^(2))/(b ^(2)) = (1)/(k) (k ne 0)` then k = ?

Options

A. 0
B. 1
C. 2
D. `1/2`

Answer

To solve the equation given in the problem, we start with the following equalities:

\[ \frac{a^2 + b^2}{c^2} = \frac{b^2 + c^2}{a^2} = \frac{c^2 + a^2}{b^2} = \frac{1}{k} \quad (k \neq 0) \]

Let's denote the common value as \( \frac{1}{k} \).

### Step 1: Set up the equations
From the first equality, we can write:
\[ \frac{a^2 + b^2}{c^2} = \frac{1}{k} \implies a^2 + b^2 = \frac{c^2}{k} \quad \text{(Equation 1)} \]

From the second equality:
\[ \frac{b^2 + c^2}{a^2} = \frac{1}{k} \implies b^2 + c^2 = \frac{a^2}{k} \quad \text{(Equation 2)} \]

From the third equality:
\[ \frac{c^2 + a^2}{b^2} = \frac{1}{k} \implies c^2 + a^2 = \frac{b^2}{k} \quad \text{(Equation 3)} \]

### Step 2: Add all three equations
Now, we will add all three equations together:
\[ (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) = \frac{c^2}{k} + \frac{a^2}{k} + \frac{b^2}{k} \]

This simplifies to:
\[ 2(a^2 + b^2 + c^2) = \frac{(a^2 + b^2 + c^2)}{k} \]

### Step 3: Factor out \( (a^2 + b^2 + c^2) \)
Assuming \( a^2 + b^2 + c^2 \neq 0 \) (which is valid since \( a, b, c \) are non-zero), we can divide both sides by \( (a^2 + b^2 + c^2) \):
\[ 2 = \frac{1}{k} \]

### Step 4: Solve for \( k \)
Rearranging gives:
\[ k = \frac{1}{2} \]

Thus, the value of \( k \) is:
\[ \boxed{2} \]