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

Options

A. a) `(2x+y)^(2)=3(x^(2)+y^(2))`
B. b) `(4x=2y)^(2)=3(x^(2)+y^(2))`
C. c) `(2x-y)^(2)=3(x^(2)+y^(2))`
D. 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 will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the given circle equation in standard form. We complete the square for \(x\) and \(y\).

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

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 these 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 has center \((-2, -1)\) and radius \(\sqrt{2}\).

### Step 2: Use the Tangent Formula
The formula for the pair of tangents from a point \((x_1, y_1)\) to a circle \((x - h)^2 + (y - k)^2 = r^2\) is given by:
\[ s \cdot s_1 = t^2 \]
where \(s\) is the equation of the circle, \(s_1\) is obtained by substituting the coordinates of the point into the circle equation, and \(t\) is the tangent equation.

### Step 3: Calculate \(s_1\)
Substituting the origin \((0, 0)\) into the circle equation:
\[ s_1 = 0^2 + 0^2 + 4(0) + 2(0) + 3 = 3 \]

### Step 4: Write the Tangent Equation
The equation of the tangents from the origin can be expressed as:
\[ x \cdot 0 + y \cdot 0 + 2x + y + 3 = 0 \]
This simplifies to:
\[ 2x + y + 3 = 0 \]

### Step 5: Substitute into the Tangent Formula
Now we substitute \(s\), \(s_1\), and \(t\) into the tangent formula:
\[ (x^2 + y^2 + 4x + 2y + 3) \cdot 3 = (2x + y + 3)^2 \]

### Step 6: Expand and Simplify
Expanding both sides:
1. Left side:
\[ 3(x^2 + y^2 + 4x + 2y + 3) = 3x^2 + 3y^2 + 12x + 6y + 9 \]
2. Right side:
\[ (2x + y + 3)^2 = 4x^2 + 4xy + y^2 + 12x + 6y + 9 \]

Setting both sides equal:
\[ 3x^2 + 3y^2 + 12x + 6y + 9 = 4x^2 + 4xy + y^2 + 12x + 6y + 9 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 0 = x^2 + xy - 2y^2 \]

### Step 8: Factor the Equation
This can be factored as:
\[ (x + 2y)(x - y) = 0 \]

### Conclusion
Thus, the pair of tangents from the origin to the circle are given by the equations:
1. \(x + 2y = 0\)
2. \(x - y = 0\)