The pair of tangents from (2,1) to the circle `x^(2)+y^(2)=4` is

Options

A. `3x^(2)+4xy+16x+8y+20=0`
B. `3x^(2)+4xy+16x-8y+20=0`
C. `3x^(2)+4xy-16x-8y+20=0`
D. `3x^(2)-4y-16x+8y-20=0`

Answer

To find the pair of tangents from the point (2, 1) to the circle given by the equation \( x^2 + y^2 = 4 \), we can follow these steps:

### Step 1: Write the equation of the circle in standard form
The equation of the circle is given as:
\[ x^2 + y^2 = 4 \]
We can rewrite this as:
\[ S: x^2 + y^2 - 4 = 0 \]

### Step 2: Identify the point from which tangents are drawn
The point from which we are drawing the tangents is \( (x_1, y_1) = (2, 1) \).

### Step 3: Calculate \( S_1 \)
We need to calculate \( S_1 \) using the coordinates of the point:
\[ S_1 = x_1^2 + y_1^2 - 4 \]
Substituting \( x_1 = 2 \) and \( y_1 = 1 \):
\[ S_1 = 2^2 + 1^2 - 4 = 4 + 1 - 4 = 1 \]

### Step 4: Use the formula for the pair of tangents
The equation of the pair of tangents from the point \( (x_1, y_1) \) to the circle is given by:
\[ SS_1 = T^2 \]
where \( S \) is the equation of the circle and \( T \) is the equation of the tangent. The equation of the tangent can be expressed as:
\[ T: x x_1 + y y_1 - S_1 = 0 \]
Substituting \( S \) and \( S_1 \):
\[ (x^2 + y^2 - 4)(1) = (2x + 1y - 1)^2 \]

### Step 5: Expand and simplify
Expanding both sides:
1. Left side: \( x^2 + y^2 - 4 \)
2. Right side:
\[ (2x + y - 1)^2 = 4x^2 + y^2 + 1 - 4xy - 4x - 2y \]
Setting these equal gives:
\[ (x^2 + y^2 - 4)(1) = 4x^2 + y^2 + 1 - 4xy - 4x - 2y \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 0 = 4x^2 + y^2 + 1 - 4xy - 4x - 2y - x^2 - y^2 + 4 \]
Combining like terms:
\[ 0 = 3x^2 - 4xy - 4x - 2y + 5 \]

### Final Equation
Thus, the equation of the pair of tangents from the point (2, 1) to the circle \( x^2 + y^2 = 4 \) is:
\[ 3x^2 + 4xy - 16x - 8y + 20 = 0 \]