The equation of the pair of straight lines parallel tox-axis and touching the circle `x^2 + y^2 – 6x – 4y - 12 = 0,` is

Options

A. `y^(2)-4y-21=0`
B. `y^(2)+4y-21=0`
C. `y^(2)-4y+21=0`
D. `y^(2)+4y+21=0`

Answer

To find the equation of the pair of straight lines that are parallel to the x-axis and touch the circle given by the equation \(x^2 + y^2 - 6x - 4y - 12 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
We start with the equation of the circle:
\[ x^2 + y^2 - 6x - 4y - 12 = 0 \]
We will rearrange this into the standard form of a circle.

### Step 2: Complete the Square
To convert the equation into standard form, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 6x \quad \text{can be rewritten as} \quad (x - 3)^2 - 9 \]
2. For \(y\):
\[ y^2 - 4y \quad \text{can be rewritten as} \quad (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 - 12 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 2)^2 - 25 = 0 \]
Thus, we have:
\[ (x - 3)^2 + (y - 2)^2 = 25 \]
This represents a circle with center \((3, 2)\) and radius \(5\).

### Step 3: Determine the Tangent Lines
Since we are looking for lines parallel to the x-axis, these lines will have the form \(y = k\). For these lines to be tangents to the circle, the distance from the center of the circle to the line must equal the radius.

### Step 4: Calculate the Distance
The distance \(d\) from the center of the circle \((3, 2)\) to the line \(y = k\) is given by:
\[ d = |2 - k| \]
Setting this equal to the radius \(5\), we have:
\[ |2 - k| = 5 \]

### Step 5: Solve for \(k\)
This absolute value equation gives us two cases:
1. \(2 - k = 5\) leads to \(k = -3\)
2. \(2 - k = -5\) leads to \(k = 7\)

Thus, the lines are \(y = -3\) and \(y = 7\).

### Step 6: Form the Equation of the Pair of Lines
The equations of the lines can be expressed as:
\[ y + 3 = 0 \quad \text{and} \quad y - 7 = 0 \]
To express this as a single equation, we can multiply these two equations:
\[ (y + 3)(y - 7) = 0 \]
Expanding this gives:
\[ y^2 - 4y - 21 = 0 \]

### Final Answer
Thus, the equation of the pair of straight lines parallel to the x-axis and touching the circle is:
\[ y^2 - 4y - 21 = 0 \]