Find the locus of a point P which is such that its polar with respect to one circle touches a second circle

Answer

To find the locus of a point \( P \) such that its polar with respect to one circle touches a second circle, we can follow these steps:

### Step 1: Define the Circles
Let the first circle be defined by the equation:
\[ x^2 + y^2 = a^2 \]
This circle is centered at the origin \( (0, 0) \) with radius \( a \).

Let the second circle be defined by the equation:
\[ (x - c)^2 + y^2 = b^2 \]
This circle is centered at \( (c, 0) \) with radius \( b \).

### Step 2: Determine the Polar of Point \( P(h, k) \)
The polar of a point \( P(h, k) \) with respect to the first circle can be expressed as:
\[ xh + yk - a^2 = 0 \]
This is the equation of the line that represents the polar of point \( P \).

### Step 3: Condition for Tangency
For the polar line to touch the second circle, the distance from the center of the second circle \( (c, 0) \) to the polar line must equal the radius \( b \) of the second circle.

The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
In our case, substituting \( A = h \), \( B = k \), and \( C = -a^2 \), the distance from the center \( (c, 0) \) to the polar line is:
\[ d = \frac{|hc + k \cdot 0 - a^2|}{\sqrt{h^2 + k^2}} = \frac{|hc - a^2|}{\sqrt{h^2 + k^2}} \]

### Step 4: Set the Distance Equal to the Radius
Setting this distance equal to the radius \( b \) of the second circle gives us:
\[ \frac{|hc - a^2|}{\sqrt{h^2 + k^2}} = b \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root yields:
\[ (hc - a^2)^2 = b^2(h^2 + k^2) \]

### Step 6: Rearranging the Equation
Expanding both sides gives:
\[ h^2c^2 - 2a^2hc + a^4 = b^2h^2 + b^2k^2 \]
Rearranging this leads to:
\[ (h^2(c^2 - b^2) - 2a^2hc + a^4 - b^2k^2 = 0 \]

### Step 7: Identify the Locus
This equation represents a conic section in the \( h, k \) plane. To express it in terms of \( x \) and \( y \), we can replace \( h \) with \( x \) and \( k \) with \( y \):
\[ x^2(c^2 - b^2) - 2a^2x + a^4 - b^2y^2 = 0 \]

This is the equation of the locus of point \( P \).

### Final Answer
The locus of point \( P \) is given by:
\[ x^2(c^2 - b^2) - 2a^2x + a^4 - b^2y^2 = 0 \]