Equation of a common tangent to the circle `x^(2)+y^(2)-6x=0` and the parabola `y^(2)=4x` is

Options

A. `2sqrt(3) y=12x+1`
B. `2sqrt(3)y=-x-12`
C. `sqrt(3)y=x+3`
D. `sqrt(3)y=3x+1`

Answer

To find the equation of a common tangent to the circle \(x^2 + y^2 - 6x = 0\) and the parabola \(y^2 = 4x\), we will follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equation of the circle in standard form. The given equation of the circle is:
\[ x^2 + y^2 - 6x = 0 \]
We can complete the square for the \(x\) terms:
\[ (x^2 - 6x) + y^2 = 0 \implies (x - 3)^2 + y^2 = 9 \]
This represents a circle with center \((3, 0)\) and radius \(3\).

The equation of the parabola is already in standard form:
\[ y^2 = 4x \]

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

### Step 3: Substitute the tangent equation into the circle's equation
Now, we substitute \(y = mx + \frac{1}{m}\) into the circle's equation:
\[ (x - 3)^2 + \left(mx + \frac{1}{m}\right)^2 = 9 \]
Expanding this gives:
\[ (x - 3)^2 + \left(m^2x^2 + 2x\cdot\frac{1}{m} + \frac{1}{m^2}\right) = 9 \]
\[ (x^2 - 6x + 9) + (m^2x^2 + \frac{2}{m}x + \frac{1}{m^2}) = 9 \]
Combining like terms:
\[ (1 + m^2)x^2 + \left(-6 + \frac{2}{m}\right)x + 9 + \frac{1}{m^2} - 9 = 0 \]
This simplifies to:
\[ (1 + m^2)x^2 + \left(-6 + \frac{2}{m}\right)x + \frac{1}{m^2} = 0 \]

### Step 4: Condition for tangency
For the tangent to touch the circle, the discriminant of the quadratic equation must be zero:
\[ b^2 - 4ac = 0 \]
Here, \(a = 1 + m^2\), \(b = -6 + \frac{2}{m}\), and \(c = \frac{1}{m^2}\).

Calculating the discriminant:
\[ \left(-6 + \frac{2}{m}\right)^2 - 4(1 + m^2)\left(\frac{1}{m^2}\right) = 0 \]

### Step 5: Solve the discriminant equation
Expanding the discriminant:
\[ (36 - 24\frac{1}{m} + \frac{4}{m^2}) - \left(\frac{4(1 + m^2)}{m^2}\right) = 0 \]
This simplifies to:
\[ 36 - 24\frac{1}{m} + \frac{4}{m^2} - \frac{4 + 4m^2}{m^2} = 0 \]
\[ 36 - 24\frac{1}{m} + \frac{4 - 4 - 4m^2}{m^2} = 0 \]
\[ 36 - 24\frac{1}{m} = 0 \]

### Step 6: Solve for \(m\)
From the above equation, we can solve for \(m\):
\[ 24\frac{1}{m} = 36 \implies m = \frac{24}{36} = \frac{2}{3} \]

### Step 7: Write the equation of the tangent line
Now substituting \(m\) back into the tangent equation:
\[ y = mx + \frac{1}{m} = \frac{2}{3}x + \frac{3}{2} \]
To write it in standard form:
\[ 2y - 3x - 3 = 0 \]

Thus, the equation of the common tangent is:
\[ 3x - 2y + 3 = 0 \]