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 condition that the lengths of the tangents from point \( P(\alpha, \beta) \) to both circles are equal. We will follow these steps:

### Step 1: Rewrite the equations of the circles
The equations of the circles are given as:
1. \( x^2 + y^2 - 4x - 5 = 0 \)
2. \( x^2 + y^2 + 6x - 2y + 6 = 0 \)

We can rewrite these equations in standard form.

**Circle 1:**
\[ x^2 - 4x + y^2 - 5 = 0 \implies (x - 2)^2 + y^2 = 9 \]
This represents a circle with center \( (2, 0) \) and radius \( 3 \).

**Circle 2:**
\[ x^2 + 6x + y^2 - 2y + 6 = 0 \implies (x + 3)^2 + (y - 1)^2 = 4 \]
This represents a circle with center \( (-3, 1) \) and radius \( 2 \).

### Step 2: Use the formula for the length of tangents
The length of the tangent from a point \( P(x_1, y_1) \) to a circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]

### Step 3: Calculate the lengths of the tangents
**For Circle 1:**
- Here, \( g = -2 \), \( f = 0 \), and \( c = -9 \).
- The length of the tangent \( L_1 \) from point \( P(\alpha, \beta) \) is:
\[ L_1 = \sqrt{\alpha^2 + \beta^2 - 4\alpha - 5} \]

**For Circle 2:**
- Here, \( g = 3 \), \( f = -1 \), and \( c = -4 \).
- The length of the tangent \( L_2 \) from point \( P(\alpha, \beta) \) is:
\[ L_2 = \sqrt{\alpha^2 + \beta^2 + 6\alpha - 2\beta + 6} \]

### Step 4: Set the lengths equal
Since the lengths of the tangents are equal, we have:
\[ L_1 = L_2 \]
Squaring both sides gives:
\[ \alpha^2 + \beta^2 - 4\alpha - 5 = \alpha^2 + \beta^2 + 6\alpha - 2\beta + 6 \]

### Step 5: Simplify the equation
Cancelling \( \alpha^2 \) and \( \beta^2 \) from both sides, we get:
\[ -4\alpha - 5 = 6\alpha - 2\beta + 6 \]
Rearranging terms leads to:
\[ -4\alpha - 6\alpha + 2\beta + 6 + 5 = 0 \]
This simplifies to:
\[ -10\alpha + 2\beta + 11 = 0 \]
Dividing through by 2 gives:
\[ 5\alpha - \beta + \frac{11}{2} = 0 \]
or equivalently:
\[ 10\alpha - 2\beta + 11 = 0 \]

### Final Result
Thus, the condition for the point \( P(\alpha, \beta) \) such that the tangents from \( P \) to both circles are equal is:
\[ 10\alpha - 2\beta + 11 = 0 \]