The range of g so that we have always a chord of contact of tangents drawn from a real point `(alpha, alpha)` to the circle `x^(2)+y^(2)+2gx+4y+2=0`, is

Options

A. (-3, 0)
B. (-4, 1)
C. (-4, 0)
D. none of these

Explanation

To find the range of \( g \) such that there is always a chord of contact of tangents drawn from the point \( (\alpha, \alpha) \) to the circle given by the equation \( x^2 + y^2 + 2gx + 4y + 2 = 0 \), we can follow these steps:

### Step 1: Identify the Circle's Center and Radius
The equation of the circle can be rewritten in standard form. The general form of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
From the given equation:
\[ x^2 + y^2 + 2gx + 4y + 2 = 0 \]
we can complete the square for \( x \) and \( y \).

For \( x \):
\[ x^2 + 2gx = (x + g)^2 - g^2 \]

For \( y \):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting these into the circle's equation gives:
\[ (x + g)^2 - g^2 + (y + 2)^2 - 4 + 2 = 0 \]
This simplifies to:
\[ (x + g)^2 + (y + 2)^2 = g^2 + 2 \]
Thus, the center of the circle is \( (-g, -2) \) and the radius \( r \) is \( \sqrt{g^2 + 2} \).

### Step 2: Determine the Condition for the Point \( (\alpha, \alpha) \)
For the point \( (\alpha, \alpha) \) to lie outside the circle, the distance from the point to the center must be greater than the radius. The distance \( d \) from the point \( (\alpha, \alpha) \) to the center \( (-g, -2) \) is given by:
\[ d = \sqrt{(\alpha + g)^2 + (\alpha + 2)^2} \]
The condition for the point to lie outside the circle is:
\[ d > r \]
Substituting the expressions for \( d \) and \( r \):
\[ \sqrt{(\alpha + g)^2 + (\alpha + 2)^2} > \sqrt{g^2 + 2} \]

### Step 3: Square Both Sides
Squaring both sides to eliminate the square roots gives:
\[ (\alpha + g)^2 + (\alpha + 2)^2 > g^2 + 2 \]
Expanding both sides:
\[ (\alpha^2 + 2\alpha g + g^2) + (\alpha^2 + 4\alpha + 4) > g^2 + 2 \]
Combining like terms:
\[ 2\alpha^2 + 2\alpha g + 4\alpha + 4 > g^2 + 2 \]
This simplifies to:
\[ 2\alpha^2 + 2\alpha g + 4\alpha + 2 > g^2 \]

### Step 4: Rearranging the Inequality
Rearranging gives:
\[ g^2 - 2\alpha g + (2\alpha^2 + 4\alpha + 2) < 0 \]
This is a quadratic inequality in \( g \).

### Step 5: Finding the Discriminant
For this quadratic to have real solutions, the discriminant must be non-negative:
\[ D = (2\alpha)^2 - 4 \cdot 1 \cdot (2\alpha^2 + 4\alpha + 2) \geq 0 \]
Calculating the discriminant:
\[ D = 4\alpha^2 - 8\alpha^2 - 16\alpha - 8 \geq 0 \]
This simplifies to:
\[ -4\alpha^2 - 16\alpha - 8 \geq 0 \]
Dividing through by -4 (and reversing the inequality):
\[ \alpha^2 + 4\alpha + 2 \leq 0 \]

### Step 6: Solving the Quadratic Inequality
The roots of the quadratic \( \alpha^2 + 4\alpha + 2 = 0 \) can be found using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \]
Thus, the roots are \( -2 + \sqrt{2} \) and \( -2 - \sqrt{2} \).

### Step 7: Finding the Range of \( g \)
The range of \( g \) can be determined from the condition \( g + 2 \) must be less than or equal to 0:
\[ g + 2 < 0 \implies g < -2 \]
And from the condition \( g + 2 \) must be greater than or equal to -4:
\[ g + 2 > -4 \implies g > -4 \]
Thus, the range of \( g \) is:
\[ -4 < g < -2 \]

### Final Answer
Therefore, the range of \( g \) such that there is always a chord of contact of tangents drawn from the point \( (\alpha, \alpha) \) to the circle is:
\[ (-4, -2) \]