If the pole of a line w.r.t to the circle `x^(2)+y^(2)=a^(2)` lies on the circle `x^(2)+y^(2)=a^(4)` then find the equation of the circle touched by the line.

Options

A. `x^(2)+y^(2)=2`
B. `x^(2)+y^(2)=1`
C. `x^(2)+y^(2)=3`
D. `x^(2)+y^(2)=4`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Identify the circles and the pole
We have two circles:
1. Circle 1: \( x^2 + y^2 = a^2 \)
2. Circle 2: \( x^2 + y^2 = a^4 \)

The pole of a line with respect to Circle 1 lies on Circle 2.

### Step 2: Determine the coordinates of the pole
Let the coordinates of the pole \( P \) on Circle 2 be given by:
\[ P(a^2 \cos \theta, a^2 \sin \theta) \]
This point lies on Circle 2, which means:
\[ (a^2 \cos \theta)^2 + (a^2 \sin \theta)^2 = a^4 \]
This simplifies to:
\[ a^4 (\cos^2 \theta + \sin^2 \theta) = a^4 \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the equation holds true.

### Step 3: Write the equation of the chord of contact
The equation of the chord of contact from point \( P \) on Circle 1 is given by:
\[ x \cos \theta + y \sin \theta = a^2 \]

### Step 4: Simplify the equation
To find the equation of the line in a simpler form, we can divide the entire equation by \( a^2 \):
\[ \frac{x \cos \theta}{a^2} + \frac{y \sin \theta}{a^2} = 1 \]
Letting \( k = \frac{1}{a^2} \), we rewrite it as:
\[ x \cos \theta + y \sin \theta = 1 \]

### Step 5: Identify the tangent line
The equation \( x \cos \theta + y \sin \theta = 1 \) represents a line that is tangent to the circle \( x^2 + y^2 = 1 \). This is because it can be viewed as the polar of the origin with respect to the circle \( x^2 + y^2 = 1 \).

### Step 6: Conclusion
Thus, the equation of the circle that is touched by the line is:
\[ x^2 + y^2 = 1 \]

### Final Answer
The equation of the circle touched by the line is:
\[ x^2 + y^2 = 1 \]

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