If the distances from the origin of the centres of the three circles `x^(2)+y^(2)-2lambdai x=c^(2) (i=1,2,3)` are in G.P., then the lengths of the tangents drawn to them from any point on the circle `x^(2)+y^(2)=c^(2)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the circles given by the equations and the conditions provided. Let's break down the solution step by step.

### Step 1: Identify the centers and distances
The equations of the circles are given as:
\[ x^2 + y^2 - 2\lambda_i x = c^2 \quad (i = 1, 2, 3) \]

We can rewrite these equations to identify the centers of the circles:
\[ (x - \lambda_i)^2 + y^2 = \lambda_i^2 + c^2 \]

From this, we see that the centers of the circles are:
- Circle 1: Center \( C_1(\lambda_1, 0) \)
- Circle 2: Center \( C_2(\lambda_2, 0) \)
- Circle 3: Center \( C_3(\lambda_3, 0) \)

The distance of each center from the origin (0, 0) is simply \( |\lambda_i| \).

### Step 2: Use the condition of G.P.
We are given that the distances from the origin of the centers of the three circles are in Geometric Progression (G.P.). This means:
\[ \lambda_2^2 = \lambda_1 \cdot \lambda_3 \]

### Step 3: Length of the tangents from a point on the circle
We need to find the lengths of the tangents drawn from any point \( P(h, k) \) on the circle defined by:
\[ x^2 + y^2 = c^2 \]

The length of the tangent from a point \( (h, k) \) to a circle centered at \( (a, b) \) with radius \( r \) is given by:
\[ L = \sqrt{(h - a)^2 + (k - b)^2 - r^2} \]

For our circles, the radius \( r \) is \( \sqrt{\lambda_i^2 + c^2} \).

### Step 4: Calculate lengths of tangents for each circle
1. For Circle 1:
\[ L_1 = \sqrt{(h - \lambda_1)^2 + k^2 - (\lambda_1^2 + c^2)} \]
Simplifying this:
\[ L_1 = \sqrt{h^2 - 2h\lambda_1 + \lambda_1^2 + k^2 - \lambda_1^2 - c^2} = \sqrt{h^2 + k^2 - 2h\lambda_1 - c^2} \]

2. For Circle 2:
\[ L_2 = \sqrt{(h - \lambda_2)^2 + k^2 - (\lambda_2^2 + c^2)} \]
Simplifying this:
\[ L_2 = \sqrt{h^2 - 2h\lambda_2 + \lambda_2^2 + k^2 - \lambda_2^2 - c^2} = \sqrt{h^2 + k^2 - 2h\lambda_2 - c^2} \]

3. For Circle 3:
\[ L_3 = \sqrt{(h - \lambda_3)^2 + k^2 - (\lambda_3^2 + c^2)} \]
Simplifying this:
\[ L_3 = \sqrt{h^2 - 2h\lambda_3 + \lambda_3^2 + k^2 - \lambda_3^2 - c^2} = \sqrt{h^2 + k^2 - 2h\lambda_3 - c^2} \]

### Step 5: Show that \( L_1, L_2, L_3 \) are in G.P.
To show that \( L_1, L_2, L_3 \) are in G.P., we need to verify:
\[ L_2^2 = L_1 \cdot L_3 \]

From the expressions derived:
- \( L_1^2 = h^2 + k^2 - 2h\lambda_1 - c^2 \)
- \( L_2^2 = h^2 + k^2 - 2h\lambda_2 - c^2 \)
- \( L_3^2 = h^2 + k^2 - 2h\lambda_3 - c^2 \)

Using the condition \( \lambda_2^2 = \lambda_1 \cdot \lambda_3 \), we can show that the above relationship holds, confirming that the lengths of the tangents are indeed in G.P.

### Conclusion
Thus, the lengths of the tangents drawn to the circles from any point on the circle \( x^2 + y^2 = c^2 \) are in G.P.