Common Tangent with Positive Slope to Circle and Hyperbola

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

1. Equations of the Given Curves

The Hyperbola:
$\frac{x^2}{9} - \frac{y^2}{4} = 1$
Here, $a^2 = 9$ and $b^2 = 4$.

The equation of any tangent to the hyperbola with slope $m$ is given by:
$y = mx \pm \sqrt{a^2 m^2 - b^2} \implies mx - y \pm \sqrt{9m^2 - 4} = 0$

The Circle:
$x^2 + y^2 - 8x = 0 \implies (x - 4)^2 + y^2 = 4^2$
The center of the circle is $C(4, 0)$ and its radius is $r = 4$.

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2. Condition of Tangency to the Circle

For the tangent line $mx - y + c = 0$ (where $c = \pm\sqrt{9m^2 - 4}$) to also be tangent to the circle, the perpendicular distance from the center $(4, 0)$ to the line must be equal to the circle's radius $r = 4$:

$\frac{|4m - 0 + c|}{\sqrt{m^2 + (-1)^2}} = 4$

Substitute $c = \pm\sqrt{9m^2 - 4}$:
$\frac{|4m \pm \sqrt{9m^2 - 4}|}{\sqrt{m^2 + 1}} = 4$

Squaring both sides:
$(4m \pm \sqrt{9m^2 - 4})^2 = 16(m^2 + 1)$
$16m^2 \pm 8m\sqrt{9m^2 - 4} + (9m^2 - 4) = 16m^2 + 16$
$\pm 8m\sqrt{9m^2 - 4} = 20 - 9m^2$

Squaring both sides again:
$64m^2(9m^2 - 4) = (20 - 9m^2)^2$
$576m^4 - 256m^2 = 400 - 360m^2 + 81m^4$
$495m^4 + 104m^2 - 400 = 0$

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3. Solving for the Slope $m$

Let $u = m^2 > 0$:
$495u^2 + 104u - 400 = 0$

Using the quadratic formula:
$u = \frac{-104 \pm \sqrt{104^2 - 4(495)(-400)}}{2 \times 495}$
$\Delta = 10816 + 792000 = 802816 = (896)^2$

Since $u = m^2 > 0$, we take the positive root:
$u = \frac{-104 + 896}{990} = \frac{792}{990} = \frac{4}{5}$

Thus:
$m^2 = \frac{4}{5}$

Since the slope is given to be positive ($m > 0$):
$m = \frac{2}{\sqrt{5}}$

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4. Finding the Intercept $c$

Recall from our tangency condition:
$8mc = 20 - 9m^2$

Substitute $m^2 = \frac{4}{5}$:
$20 - 9\left(\frac{4}{5}\right) = 20 - \frac{36}{5} = \frac{64}{5} > 0$

Since $m > 0$ and the right-hand side is positive, we must have $c > 0$:
$c = +\sqrt{9m^2 - 4} = \sqrt{9\left(\frac{4}{5}\right) - 4} = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}}$

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5. Equation of the Common Tangent

Substituting $m = \frac{2}{\sqrt{5}}$ and $c = \frac{4}{\sqrt{5}}$ into $y = mx + c$:
$y = \frac{2}{\sqrt{5}}x + \frac{4}{\sqrt{5}}$

Multiplying through by $\sqrt{5}$:
$\sqrt{5}y = 2x + 4 \implies 2x - \sqrt{5}y + 4 = 0$

This matches Option B.