The locus of poles of tangents to the circle `(x-p)^(2)+y^(2)=b^(2)` w.r.t. the circle `x^(2)+y^(2)=a^(2)` is

Options

A. `(a^(2)-px)^(2)=b^(2)(x^(2)+y^(2))`
B. `(a^(2)_bx)^(2)=p^(2)(x^(2)+y^(2))`
C. `(a^(2)+px)^(2)=b^(2)(x^(2)+y^(2))`
D. `(a^(2)+bx)^(2)=p^(2)(x^(2)+y^(2))`

Answer

To find the locus of poles of tangents to the circle \((x - p)^2 + y^2 = b^2\) with respect to the circle \(x^2 + y^2 = a^2\), we can follow these steps:

### Step 1: Understand the given circles
The first circle has the equation \((x - p)^2 + y^2 = b^2\), which is centered at \((p, 0)\) with a radius of \(b\). The second circle has the equation \(x^2 + y^2 = a^2\), which is centered at the origin \((0, 0)\) with a radius of \(a\).

### Step 2: Write the equation of the tangent to the first circle
The general equation of the tangent to the circle \((x - p)^2 + y^2 = b^2\) at the point \((x_1, y_1)\) on the circle is given by:
\[ (x_1 - p)(x - p) + y_1 y = b^2 \]
This can be rearranged to:
\[ x(x_1 - p) + yy_1 = b^2 + p(x_1 - p) \]

### Step 3: Write the equation of the polar of a point with respect to the second circle
For a point \((x_2, y_2)\) on the second circle \(x^2 + y^2 = a^2\), the polar of this point with respect to the circle is given by:
\[ xx_2 + yy_2 = a^2 \]

### Step 4: Set up the ratio of the coefficients
Since the tangents from the first circle and the polar from the second circle must be proportional, we can write:
\[ \frac{x_1 - p}{x_2} = \frac{y_1}{y_2} = \frac{b^2 + p(x_1 - p)}{a^2} \]

### Step 5: Solve for \(x_1 - p\) and \(y_1\)
From the first ratio:
\[ x_1 - p = \frac{b^2 x_2}{a^2 - p x_2} \]
From the second ratio:
\[ y_1 = \frac{x_1 - p}{x_2} y_2 \]

Substituting the expression for \(x_1 - p\) into the equation for \(y_1\):
\[ y_1 = \frac{b^2}{a^2 - p x_2} y_2 \]

### Step 6: Substitute into the circle equation
Since the point \((x_1, y_1)\) lies on the first circle, we have:
\[ (x_1 - p)^2 + y_1^2 = b^2 \]
Substituting the expressions for \(x_1 - p\) and \(y_1\):
\[ \left(\frac{b^2 x_2}{a^2 - p x_2}\right)^2 + \left(\frac{b^2 y_2}{a^2 - p x_2}\right)^2 = b^2 \]

### Step 7: Simplify the equation
This simplifies to:
\[ \frac{b^4 (x_2^2 + y_2^2)}{(a^2 - p x_2)^2} = b^2 \]
Since \(x_2^2 + y_2^2 = a^2\) (as \((x_2, y_2)\) lies on the second circle), we can substitute:
\[ \frac{b^4 a^2}{(a^2 - p x_2)^2} = b^2 \]
Cross-multiplying gives:
\[ b^4 a^2 = b^2 (a^2 - p x_2)^2 \]
Dividing both sides by \(b^2\) (assuming \(b \neq 0\)):
\[ b^2 a^2 = (a^2 - p x_2)^2 \]

### Step 8: Final form of the locus
Taking the square root gives:
\[ \pm ba = a^2 - p x_2 \]
Thus, we can express \(x_2\) in terms of \(y_2\) and \(p\), leading to the final equation of the locus.

### Conclusion
The locus of the poles of the tangents to the circle \((x - p)^2 + y^2 = b^2\) with respect to the circle \(x^2 + y^2 = a^2\) is given by:
\[ b^2 x^2 + y^2 = a^2 - p^2 \]