If the square of the length of the tangents from a point P to the circles `x^(2)+y^(2)=a^(2), x^(2)+y^(2)=b^(2), x^(2)+y^(2)=c^(2)` are in A.P. then `a^(2),b^(2),c^(2)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. A.G.P.

Answer

To solve the problem, we need to show that if the squares of the lengths of the tangents from a point \( P \) to the circles \( x^2 + y^2 = a^2 \), \( x^2 + y^2 = b^2 \), and \( x^2 + y^2 = c^2 \) are in arithmetic progression (A.P.), then \( a^2, b^2, c^2 \) are also in A.P.

### Step-by-Step Solution:

1. **Identify the point and the circles**: Let the point \( P \) be \( (h, k) \). The equations of the circles are:
- 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. **Calculate the lengths of the tangents**:
The length of the tangent from the point \( P(h, k) \) to a circle \( x^2 + y^2 = r^2 \) is given by the formula:
\[ L = \sqrt{h^2 + k^2 - r^2} \]
Therefore, we have:
- Length of tangent to Circle 1:
\[ L_1 = \sqrt{h^2 + k^2 - a^2} \]
- Length of tangent to Circle 2:
\[ L_2 = \sqrt{h^2 + k^2 - b^2} \]
- Length of tangent to Circle 3:
\[ L_3 = \sqrt{h^2 + k^2 - c^2} \]

3. **Square the lengths of the tangents**:
Now, squaring the lengths gives:
\[ L_1^2 = h^2 + k^2 - a^2 \]
\[ L_2^2 = h^2 + k^2 - b^2 \]
\[ L_3^2 = h^2 + k^2 - c^2 \]

4. **Set up the condition for A.P.**:
According to the problem, \( L_1^2, L_2^2, L_3^2 \) are in A.P. This means:
\[ 2L_2^2 = L_1^2 + L_3^2 \]
Substituting the squared lengths:
\[ 2(h^2 + k^2 - b^2) = (h^2 + k^2 - a^2) + (h^2 + k^2 - c^2) \]

5. **Simplify the equation**:
Expanding and simplifying gives:
\[ 2h^2 + 2k^2 - 2b^2 = 2h^2 + 2k^2 - a^2 - c^2 \]
Canceling \( 2h^2 + 2k^2 \) from both sides:
\[ -2b^2 = -a^2 - c^2 \]
Rearranging gives:
\[ a^2 + c^2 = 2b^2 \]

6. **Conclusion**:
The equation \( a^2 + c^2 = 2b^2 \) shows that \( a^2, b^2, c^2 \) are in arithmetic progression (A.P.).

### Final Result:
Thus, we conclude that if the squares of the lengths of the tangents from point \( P \) to the circles are in A.P., then \( a^2, b^2, c^2 \) are also in A.P.

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