Find the pair of tangents from origin to `x^(2)+y^(2)+4x+2y+3=0`

Options

A. `(2x+y)^(2)=3(x^(2)+y^(2))`
B. `(2x+y)^(2)=33(x^(2)+y^(2))`
C. `(2x-y)^(2)=-3(x^(2)-y^(2))`
D. not existing

Answer

To find the pair of tangents from the origin to the circle given by the equation \( x^2 + y^2 + 4x + 2y + 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. We can complete the square for both \( x \) and \( y \).

The given equation is:
\[ x^2 + y^2 + 4x + 2y + 3 = 0 \]

Rearranging gives:
\[ x^2 + 4x + y^2 + 2y + 3 = 0 \]

Completing the square for \( x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

Completing the square for \( y \):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting back into the equation:
\[ (x + 2)^2 - 4 + (y + 1)^2 - 1 + 3 = 0 \]
\[ (x + 2)^2 + (y + 1)^2 - 2 = 0 \]
\[ (x + 2)^2 + (y + 1)^2 = 2 \]

This shows that the circle is centered at \( (-2, -1) \) with a radius of \( \sqrt{2} \).

### Step 2: Identify the Point and Circle Parameters
The center of the circle is \( (h, k) = (-2, -1) \) and the radius \( r = \sqrt{2} \). The point from which we are drawing tangents is the origin \( (0, 0) \).

### Step 3: Use the Tangent Formula
The formula for the tangents from a point \( (x_1, y_1) \) to the circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by:
\[ T^2 = S_1 \]
where \( S = x^2 + y^2 + 2hx + 2ky + c \) and \( S_1 \) is obtained by substituting \( (x_1, y_1) \).

Here, \( S \) for our circle is:
\[ S = x^2 + y^2 + 4x + 2y + 3 \]
Substituting \( (x_1, y_1) = (0, 0) \):
\[ S_1 = 0^2 + 0^2 + 4(0) + 2(0) + 3 = 3 \]

### Step 4: Write the Tangent Equation
The equation of the tangent from the origin can be expressed as:
\[ T = xx_1 + yy_1 + hx + ky + c = 0 \]
Substituting \( (x_1, y_1) = (0, 0) \):
\[ T = 0 + 0 + 2x + y + 3 = 2x + y + 3 \]

### Step 5: Find the Pair of Tangents
Now we can find the pair of tangents using:
\[ S \cdot S_1 = T^2 \]
Substituting \( S \) and \( S_1 \):
\[ (x^2 + y^2 + 4x + 2y + 3) \cdot 3 = (2x + y + 3)^2 \]

Expanding both sides:
\[ 3(x^2 + y^2 + 4x + 2y + 3) = 4x^2 + 4xy + y^2 + 12x + 6y + 9 \]

This leads to:
\[ 3x^2 + 3y^2 + 12x + 6y + 9 = 4x^2 + 4xy + y^2 + 12x + 6y + 9 \]

Simplifying gives:
\[ 0 = x^2 + 4xy - 2y^2 \]

### Step 6: Factor the Quadratic
Factoring the quadratic:
\[ x^2 + 4xy - 2y^2 = 0 \]

This can be solved to find the slopes of the tangents.

### Conclusion
Thus, the pair of tangents from the origin to the circle \( x^2 + y^2 + 4x + 2y + 3 = 0 \) can be derived from the above equations.