Polar of origin (0, 0) w.r.t. the circle `x^(2)+y^(2)+2lambda x +2 mu y+c=0` touches the circle `x^(2)+y^(2)=r^(2)` if

Options

A. `c=r(lambda^(2)+mu^(2))`
B. `r=c(lambda^(2)+mu^(2))`
C. `c^(2)=r^(2)(lambda^(2)+mu^(2))`
D. `r^(2)=c^(2)(lambda^(2)+mu^(2))`

Answer

To determine the condition under which the polar of the origin (0, 0) with respect to the circle \(x^2 + y^2 + 2\lambda x + 2\mu y + c = 0\) touches the circle \(x^2 + y^2 = r^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given circles**:
- The first circle is given by the equation:
\[ S_1: x^2 + y^2 + 2\lambda x + 2\mu y + c = 0 \]
- The second circle is:
\[ S_2: x^2 + y^2 = r^2 \]

2. **Find the polar of the origin with respect to the first circle**:
- The polar of a point \((x_0, y_0)\) with respect to the circle \(S_1 = 0\) is given by:
\[ S_1(x_0, y_0) = 0 \]
- For the origin \((0, 0)\), substituting \(x_0 = 0\) and \(y_0 = 0\) into the equation gives:
\[ \lambda x + \mu y + c = 0 \]
- Thus, the equation of the polar line is:
\[ \lambda x + \mu y + c = 0 \]

3. **Determine the condition for tangency**:
- The polar line touches the second circle \(S_2\) if the perpendicular distance from the center of the second circle (which is at the origin \((0, 0)\)) to the line \(\lambda x + \mu y + c = 0\) is equal to the radius \(r\).
- The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \(A = \lambda\), \(B = \mu\), and \(C = c\). Thus, the distance from the origin to the line is:
\[ d = \frac{|\lambda(0) + \mu(0) + c|}{\sqrt{\lambda^2 + \mu^2}} = \frac{|c|}{\sqrt{\lambda^2 + \mu^2}} \]

4. **Set the distance equal to the radius**:
- For the polar to touch the circle, this distance must equal the radius \(r\):
\[ \frac{|c|}{\sqrt{\lambda^2 + \mu^2}} = r \]

5. **Square both sides to eliminate the square root**:
- Squaring both sides gives:
\[ c^2 = r^2(\lambda^2 + \mu^2) \]

### Final Condition:
Thus, the condition for the polar of the origin to touch the circle \(x^2 + y^2 = r^2\) is:
\[ c^2 = r^2(\lambda^2 + \mu^2) \]