Find the value of p so that the straight line ` x cos alpha + y sin alpha - p = 0 ` may touch the circle
` x^(2) + y^(2) - 2a x cos alpha - 2by sin alpha - a^(2) sin ^(2) alpha = 0`

Answer

To find the value of \( p \) such that the line \( x \cos \alpha + y \sin \alpha - p = 0 \) touches the circle given by

\[ x^2 + y^2 - 2ax \cos \alpha - 2by \sin \alpha - a^2 \sin^2 \alpha = 0, \]

we can follow these steps:

### Step 1: Identify the center and radius of the circle

The general form of the circle is:

\[ x^2 + y^2 - 2hx - 2ky + c = 0, \]

where \( (h, k) \) is the center and \( r \) is the radius.

From the given circle equation, we can rewrite it as:

\[ x^2 + y^2 - 2a \cos \alpha \cdot x - 2b \sin \alpha \cdot y - a^2 \sin^2 \alpha = 0. \]

Thus, the center \( (h, k) \) is:

\[ h = a \cos \alpha, \quad k = b \sin \alpha. \]

### Step 2: Calculate the radius of the circle

The radius \( r \) can be found using the formula:

\[ r = \sqrt{h^2 + k^2 - c}. \]

Here, \( c = -a^2 \sin^2 \alpha \), so:

\[ r = \sqrt{(a \cos \alpha)^2 + (b \sin \alpha)^2 + a^2 \sin^2 \alpha}. \]

This simplifies to:

\[ r = \sqrt{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + a^2 \sin^2 \alpha} = \sqrt{a^2 (\cos^2 \alpha + \sin^2 \alpha) + b^2 \sin^2 \alpha} = \sqrt{a^2 + b^2 \sin^2 \alpha}. \]

### Step 3: Find the distance from the center to the line

The distance \( d \) from the center \( (h, k) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}, \]

where \( (x_0, y_0) \) is the center of the circle.

For our line \( x \cos \alpha + y \sin \alpha - p = 0 \), we have:

- \( A = \cos \alpha \)
- \( B = \sin \alpha \)
- \( C = -p \)

Thus, the distance \( d \) becomes:

\[ d = \frac{|a \cos^2 \alpha + b \sin^2 \alpha - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |a \cos^2 \alpha + b \sin^2 \alpha - p|. \]

### Step 4: Set the distance equal to the radius

Since the line touches the circle, the distance \( d \) must equal the radius \( r \):

\[ |a \cos^2 \alpha + b \sin^2 \alpha - p| = \sqrt{a^2 + b^2 \sin^2 \alpha}. \]

### Step 5: Solve for \( p \)

This gives us two cases to consider:

1. \( a \cos^2 \alpha + b \sin^2 \alpha - p = \sqrt{a^2 + b^2 \sin^2 \alpha} \)
2. \( a \cos^2 \alpha + b \sin^2 \alpha - p = -\sqrt{a^2 + b^2 \sin^2 \alpha} \)

From the first case:

\[ p = a \cos^2 \alpha + b \sin^2 \alpha - \sqrt{a^2 + b^2 \sin^2 \alpha}. \]

From the second case:

\[ p = a \cos^2 \alpha + b \sin^2 \alpha + \sqrt{a^2 + b^2 \sin^2 \alpha}. \]

Thus, the values of \( p \) that allow the line to touch the circle are:

\[ p = a \cos^2 \alpha + b \sin^2 \alpha \pm \sqrt{a^2 + b^2 \sin^2 \alpha}. \]