The locus of the poles of the line `ax+by+c=0` w.r.t a system of circles `x^(2)+y^(2)=lamda` where `lamda` is parameter is

Options

A. `ax+by=lamda`
B. `bx+ay=lamda`
C. `ax-by=0`
D. `bx-ay=0`

Answer

To find the locus of the poles of the line \( ax + by + c = 0 \) with respect to the system of circles \( x^2 + y^2 = \lambda \), we can follow these steps:

### Step 1: Understand the given equations
We have a line given by the equation:
\[ ax + by + c = 0 \]
And a family of circles given by:
\[ x^2 + y^2 = \lambda \]
where \( \lambda \) is a parameter.

### Step 2: Define the pole of the line
Let the pole of the line \( ax + by + c = 0 \) with respect to the circle \( x^2 + y^2 = \lambda \) be \( (h, k) \).

### Step 3: Use the polar equation
The polar of the point \( (h, k) \) with respect to the circle \( x^2 + y^2 = \lambda \) is given by:
\[ hx + ky = \lambda \]
This is the equation of the polar line corresponding to the point \( (h, k) \).

### Step 4: Set the polar equal to the line equation
Since the line \( ax + by + c = 0 \) and the polar line \( hx + ky = \lambda \) represent the same line, we can equate them:
\[ hx + ky = \lambda \quad \text{(1)} \]
\[ ax + by + c = 0 \quad \text{(2)} \]

### Step 5: Rearranging the equations
From equation (2), we can express \( y \) in terms of \( x \):
\[ by = -ax - c \quad \Rightarrow \quad y = -\frac{a}{b}x - \frac{c}{b} \]

### Step 6: Substitute into the polar equation
Substituting \( y \) from equation (2) into equation (1):
\[ hx + k\left(-\frac{a}{b}x - \frac{c}{b}\right) = \lambda \]
This simplifies to:
\[ hx - \frac{ak}{b}x - \frac{kc}{b} = \lambda \]

### Step 7: Combine like terms
Combining the \( x \) terms gives:
\[ \left(h - \frac{ak}{b}\right)x - \frac{kc}{b} = \lambda \]

### Step 8: Set the coefficients equal
For this equation to hold for all \( x \), the coefficients must be equal. Thus, we have:
1. \( h - \frac{ak}{b} = 0 \)
2. \( -\frac{kc}{b} = \lambda \)

From the first equation:
\[ h = \frac{ak}{b} \]

### Step 9: Express \( k \) in terms of \( h \)
Substituting \( h \) back into the second equation gives:
\[ -\frac{kc}{b} = \lambda \]
Substituting \( k = \frac{bh}{a} \) into this equation leads to:
\[ -\frac{\left(\frac{bh}{a}\right)c}{b} = \lambda \quad \Rightarrow \quad -\frac{hc}{a} = \lambda \]

### Step 10: Find the locus
From the relationships established, we can derive the locus of the poles:
\[ b k = -a h \quad \Rightarrow \quad b y = -a x \]
This represents the line:
\[ bx + ay = 0 \]

### Final Result
Thus, the locus of the poles of the line \( ax + by + c = 0 \) with respect to the system of circles \( x^2 + y^2 = \lambda \) is given by:
\[ bx + ay = 0 \]

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