The equation of the common tangent touching the circle `(x-3)^(2)+y^(2)=9` and the parabola `y^(2)=4x` below the x-axis is

Options

A. `sqrt3y=3x+1`
B. `sqrt3y=-(x+3)`
C. `sqrt3y=x+3`
D. `sqrt3y=-(3x+1)`

Answer

To find the equation of the common tangent touching the circle \((x-3)^2 + y^2 = 9\) and the parabola \(y^2 = 4x\) below the x-axis, we can follow these steps:

### Step 1: Identify the Circle and Parabola
The given circle has its center at \((3, 0)\) and a radius of \(3\) (since \(9 = 3^2\)). The parabola \(y^2 = 4x\) opens to the right.

### Step 2: Write the General Equation of the Tangent to the Parabola
The equation of the tangent to the parabola \(y^2 = 4x\) can be expressed as:
\[ y = mx + \frac{1}{m} \]
where \(m\) is the slope of the tangent.

### Step 3: Find the Point of Tangency with the Circle
The tangent must also touch the circle. The equation of the tangent line can be written in the form:
\[ y = mx + c \]
where \(c = \frac{1}{m}\).

### Step 4: Substitute the Tangent Line into the Circle's Equation
Substituting \(y = mx + c\) into the circle's equation:
\[ (x - 3)^2 + (mx + c)^2 = 9 \]
Expanding this gives:
\[ (x - 3)^2 + (mx + \frac{1}{m})^2 = 9 \]
\[ (x - 3)^2 + (m^2x^2 + 2mx\frac{1}{m} + \frac{1}{m^2}) = 9 \]

### Step 5: Rearranging the Equation
Rearranging the equation leads to a quadratic in \(x\):
\[ (1 + m^2)x^2 + (2m\frac{1}{m} - 6)x + (9 - 9) = 0 \]
This simplifies to:
\[ (1 + m^2)x^2 + (2 - 6m)x = 0 \]

### Step 6: Condition for Tangency
For the line to be a tangent, the discriminant of this quadratic must be zero:
\[ (2 - 6m)^2 - 4(1 + m^2)(0) = 0 \]
This leads to:
\[ (2 - 6m)^2 = 0 \]
Thus:
\[ 2 - 6m = 0 \implies m = \frac{1}{3} \]

### Step 7: Find the Value of \(c\)
Substituting \(m = \frac{1}{3}\) into \(c = \frac{1}{m}\):
\[ c = 3 \]

### Step 8: Write the Equation of the Tangent
The equation of the tangent line is:
\[ y = \frac{1}{3}x + 3 \]

### Step 9: Adjust for the Tangent Below the x-axis
Since we need the tangent below the x-axis, we take the negative slope:
\[ y = -\frac{1}{3}x - 3 \]

### Final Answer
Thus, the equation of the common tangent touching the circle and the parabola below the x-axis is:
\[ y = -\frac{1}{3}x - 3 \]

---