Given the circles `x^(2)+y^(2)-4x-5=0` and `x^(2)+y^(2)+6x-2y+6=0`.
Let P be a point `(alpha,beta)` such that the tangents from P to both the circles are equal. Then

Options

A. `2alpha+10beta+11=0`
B. `2alpha-10beta+11=0`
C. `10alpha-2beta+11=0`
D. `10alpha+2beta+11=0`

Answer

To solve the problem, we need to find the conditions under which the lengths of the tangents drawn from a point \( P(\alpha, \beta) \) to both given circles are equal.

### Step-by-Step Solution:

1. **Identify the equations of the circles:**
The first circle is given by:
\[ x^2 + y^2 - 4x - 5 = 0 \]
The second circle is given by:
\[ x^2 + y^2 + 6x - 2y + 6 = 0 \]

2. **Rewrite the equations in standard form:**
For the first circle:
\[ (x^2 - 4x) + y^2 = 5 \implies (x - 2)^2 + y^2 = 9 \]
This circle has center \( (2, 0) \) and radius \( 3 \).

For the second circle:
\[ (x^2 + 6x) + (y^2 + 2y) = -6 \implies (x + 3)^2 + (y + 1)^2 = 4 \]
This circle has center \( (-3, -1) \) and radius \( 2 \).

3. **Use the formula for the length of tangents:**
The length of the tangent from point \( P(\alpha, \beta) \) to a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ L = \sqrt{(\alpha - h)^2 + (\beta - k)^2 - r^2} \]

4. **Calculate the lengths of the tangents to both circles:**
For the first circle \( S_1 \):
\[ L_1 = \sqrt{(\alpha - 2)^2 + (\beta - 0)^2 - 3^2} = \sqrt{(\alpha - 2)^2 + \beta^2 - 9} \]

For the second circle \( S_2 \):
\[ L_2 = \sqrt{(\alpha + 3)^2 + (\beta + 1)^2 - 2^2} = \sqrt{(\alpha + 3)^2 + (\beta + 1)^2 - 4} \]

5. **Set the lengths equal:**
Since the lengths of the tangents are equal, we have:
\[ L_1 = L_2 \]
Squaring both sides gives:
\[ (\alpha - 2)^2 + \beta^2 - 9 = (\alpha + 3)^2 + (\beta + 1)^2 - 4 \]

6. **Expand and simplify the equation:**
Expanding both sides:
\[ (\alpha^2 - 4\alpha + 4 + \beta^2 - 9) = (\alpha^2 + 6\alpha + 9 + \beta^2 + 2\beta + 1 - 4) \]
Simplifying:
\[ \alpha^2 - 4\alpha + \beta^2 - 5 = \alpha^2 + 6\alpha + \beta^2 + 2\beta + 6 \]
Canceling \( \alpha^2 \) and \( \beta^2 \) from both sides:
\[ -4\alpha - 5 = 6\alpha + 2\beta + 6 \]

7. **Rearranging the equation:**
Bringing all terms to one side:
\[ -4\alpha - 6\alpha - 2\beta - 5 - 6 = 0 \]
Simplifying:
\[ -10\alpha - 2\beta - 11 = 0 \]
Dividing by -1:
\[ 10\alpha + 2\beta + 11 = 0 \]

8. **Final equation:**
Rearranging gives:
\[ 10\alpha + 2\beta + 11 = 0 \]