The chord of contact of tangents drawn from any point on the circle `x^(2)+y^(2)=a^(2)` to `x^(2)+y^(2)=b^(2)`, touches the circle `x^(2)+y^(2)=c^(2), a gt b` then a, b, c are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None

Answer

To solve the problem, we need to analyze the conditions given for the circles and the chord of contact.

### Step-by-Step Solution:

1. **Identify the Circles:**
We have three circles:
- Circle 1: \( x^2 + y^2 = a^2 \)
- Circle 2: \( x^2 + y^2 = b^2 \)
- Circle 3: \( x^2 + y^2 = c^2 \)

2. **Point on Circle 1:**
Let a point \( P(x_1, y_1) \) be on Circle 1. Therefore, it satisfies the equation:
\[ x_1^2 + y_1^2 = a^2 \]

3. **Chord of Contact for Circle 2:**
The equation of the chord of contact from point \( P(x_1, y_1) \) to Circle 2 is given by:
\[ xx_1 + yy_1 = b^2 \]

4. **Condition for Tangent to Circle 3:**
This chord of contact must also be a tangent to Circle 3. The distance from the center of Circle 3 (which is at the origin (0,0)) to the line \( xx_1 + yy_1 = b^2 \) must equal the radius of Circle 3, which is \( c \).

5. **Finding the Distance from the Origin to the Line:**
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( xx_1 + yy_1 - b^2 = 0 \), we have:
- \( A = x_1 \)
- \( B = y_1 \)
- \( C = -b^2 \)

Thus, the distance from the origin (0,0) to the line is:
\[ d = \frac{|b^2|}{\sqrt{x_1^2 + y_1^2}} = \frac{b^2}{\sqrt{a^2}} = \frac{b^2}{a} \]

6. **Setting the Distance Equal to the Radius:**
Since this distance must equal the radius \( c \) of Circle 3, we have:
\[ \frac{b^2}{a} = c \]

7. **Rearranging the Equation:**
Rearranging gives us:
\[ b^2 = ac \]

8. **Concluding the Relationship:**
Since \( a > b \), we can conclude that \( a, b, c \) are in geometric progression (GP). This is because the relationship \( b^2 = ac \) is the condition for three numbers to be in GP.

### Final Result:
Thus, we conclude that \( a, b, c \) are in GP.