If the lines given by `ax^(2)+2hxy+by^(2)=0` are equally inclined to the lines given by `ax^(2)+2hxy+by^(2)+lambda(x^(2)+y^(2))=0`, then

Answer

To solve the problem, we need to analyze the conditions under which the lines represented by the equations are equally inclined to each other. We will derive the necessary conditions step by step.

### Step 1: Understand the given equations
We have two equations of lines:
1. \( ax^2 + 2hxy + by^2 = 0 \) (Equation 1)
2. \( ax^2 + 2hxy + by^2 + \lambda(x^2 + y^2) = 0 \) (Equation 2)

### Step 2: Rewrite the second equation
We can rewrite Equation 2 as:
\[ (ax + \lambda)x^2 + 2hxy + (b + \lambda)y^2 = 0 \]
This shows that the second equation is a linear combination of the first equation with an additional term involving \(\lambda\).

### Step 3: Find the angle bisector of the first equation
The angle bisector of the lines given by Equation 1 can be derived from the coefficients:
\[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \]
This is the equation of the angle bisector for the first set of lines.

### Step 4: Find the angle bisector of the second equation
For the second equation, the angle bisector can be derived similarly:
\[ \frac{x^2 - y^2}{(a + \lambda) - (b + \lambda)} = \frac{xy}{h} \]
This simplifies to:
\[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \]

### Step 5: Compare the angle bisectors
From the angle bisector equations derived from both equations, we see that:
- The angle bisector of Equation 1 is \(\frac{x^2 - y^2}{a - b} = \frac{xy}{h}\)
- The angle bisector of Equation 2 is also \(\frac{x^2 - y^2}{a - b} = \frac{xy}{h}\)

Since both angle bisectors are identical, this means that the lines represented by both equations are equally inclined to each other.

### Step 6: Conclusion about \(\lambda\)
Since the angle bisectors are the same for any value of \(\lambda\), we conclude that the lines are equally inclined for any real number value of \(\lambda\).

### Final Answer
Thus, the value of \(\lambda\) can be any real number.

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