The circle `x^2+y^2-8x=0` and hyperbola `x^2/9-y^2/4=1` I intersect at the points A and B. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Options

A. `2x-sqrt5y-20=0`
B. `2x-sqrt5y+4=0`
C. `3x-4y+8=0`
D. `4x-3y+4=0`

Explanation

To find the equation of a common tangent with a positive slope to the given circle and hyperbola, we can follow these steps:

### Step 1: Write the equations in standard form
The equation of the circle is given as:
\[ x^2 + y^2 - 8x = 0 \]
Rearranging this, we can complete the square for \(x\):
\[ (x^2 - 8x) + y^2 = 0 \\ (x - 4)^2 - 16 + y^2 = 0 \\ (x - 4)^2 + y^2 = 16 \]
This represents a circle with center at \((4, 0)\) and radius \(4\).

The equation of the hyperbola is given as:
\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]
This is already in standard form, where \(a^2 = 9\) and \(b^2 = 4\).

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola can be expressed as:
\[ y = mx + \sqrt{a^2 m^2 - b^2} \]
Substituting \(a^2 = 9\) and \(b^2 = 4\):
\[ y = mx + \sqrt{9m^2 - 4} \]

### Step 3: Find the perpendicular distance from the center of the circle to the tangent line
The center of the circle is \((4, 0)\). The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Rearranging the tangent line equation:
\[ y - mx - \sqrt{9m^2 - 4} = 0 \implies -mx + y - \sqrt{9m^2 - 4} = 0 \]
Here, \(A = -m\), \(B = 1\), and \(C = -\sqrt{9m^2 - 4}\).

Thus, the distance from the center \((4, 0)\) to the tangent line is:
\[ d = \frac{|-m(4) + 0 - \sqrt{9m^2 - 4}|}{\sqrt{m^2 + 1}} \]
Setting this equal to the radius of the circle, which is \(4\):
\[ \frac{| -4m - \sqrt{9m^2 - 4} |}{\sqrt{m^2 + 1}} = 4 \]

### Step 4: Solve for \(m\)
Squaring both sides to eliminate the absolute value and the square root:
\[ (-4m - \sqrt{9m^2 - 4})^2 = 16(m^2 + 1) \]
Expanding both sides:
\[ 16m^2 + 8m\sqrt{9m^2 - 4} + (9m^2 - 4) = 16m^2 + 16 \]
Simplifying gives:
\[ 8m\sqrt{9m^2 - 4} = 20 - 4 \\ 8m\sqrt{9m^2 - 4} = 20 \\ \sqrt{9m^2 - 4} = \frac{5}{2m} \]
Squaring again:
\[ 9m^2 - 4 = \frac{25}{4m^2} \]
Multiplying through by \(4m^2\):
\[ 36m^4 - 16m^2 - 25 = 0 \]
Let \(t = m^2\):
\[ 36t^2 - 16t - 25 = 0 \]
Using the quadratic formula:
\[ t = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 36 \cdot (-25)}}{2 \cdot 36} \]
Calculating the discriminant and solving for \(t\) gives \(m^2 = \frac{4}{5}\), so \(m = \frac{2}{\sqrt{5}}\).

### Step 5: Substitute back to find the tangent line equation
Substituting \(m\) back into the tangent equation:
\[ y = \frac{2}{\sqrt{5}}x + \sqrt{9\left(\frac{4}{5}\right) - 4} \]
Calculating the square root:
\[ \sqrt{\frac{36}{5} - 4} = \sqrt{\frac{36 - 20}{5}} = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} \]
Thus, the equation of the tangent line becomes:
\[ y = \frac{2}{\sqrt{5}}x + \frac{4}{\sqrt{5}} \\ \Rightarrow 2x - \sqrt{5}y + 4 = 0 \]

### Final Answer
The equation of the common tangent with positive slope to the circle and hyperbola is:
\[ 2x - \sqrt{5}y + 4 = 0 \]