If the polar of a point `(p,q)` with respect to the circle `x^2 +y^2=a^2` touches the circle `(x-c)^2 + (y-d)^2 =b^2`, then

Options

A. `b^(2) ( p^(2) + q^(2)) = ( a^(2) -cp -dp)^(2)`
B. `b^(2) ( p^(2) + q^(2)) = ( a^(2) -cq -dp)^(2)`
C. `a^(2) ( p^(2) + q^(2)) = ( b^(2) -cp -dp)^(2)`
D. None of these

Answer

To solve the problem, we need to find the condition under which the polar of the point \((p, q)\) with respect to the circle \(x^2 + y^2 = a^2\) touches the circle \((x - c)^2 + (y - d)^2 = b^2\).

### Step-by-Step Solution:

1. **Identify the Polar Equation:**
The polar of a point \((p, q)\) with respect to the circle \(x^2 + y^2 = a^2\) is given by the equation:
\[ px + qy - a^2 = 0 \]

2. **Identify the Circle Equation:**
The equation of the second circle is:
\[ (x - c)^2 + (y - d)^2 = b^2 \]

3. **Determine the Center and Radius of the Second Circle:**
The center of the circle \((x - c)^2 + (y - d)^2 = b^2\) is \((c, d)\) and its radius is \(b\).

4. **Calculate the Perpendicular Distance from the Center of the Circle to the Polar Line:**
The distance \(D\) from the center \((c, d)\) of the second circle to the line \(px + qy - a^2 = 0\) is given by the formula:
\[ D = \frac{|pc + qd - a^2|}{\sqrt{p^2 + q^2}} \]

5. **Set the Condition for Tangency:**
For the polar line to touch the second circle, the distance \(D\) must be equal to the radius \(b\):
\[ \frac{|pc + qd - a^2|}{\sqrt{p^2 + q^2}} = b \]

6. **Square Both Sides to Eliminate the Absolute Value:**
Squaring both sides gives:
\[ (pc + qd - a^2)^2 = b^2(p^2 + q^2) \]

This equation represents the condition under which the polar of the point \((p, q)\) with respect to the first circle touches the second circle.

### Final Result:
The condition is:
\[ (pc + qd - a^2)^2 = b^2(p^2 + q^2) \]