Equation of the common tangent, with positive slope, to the circle \(x^2 + y^2 - 8x = 0\) as well as to the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\), is:
Step-by-step Solution:
The equation of a tangent of slope m to the hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \) is \[ y = mx + \sqrt{9m^2 - 4} \] If it touches the circle \(x^2 + y^2 - 8x = 0\), then \[ \left| \frac{4m + \sqrt{9m^2 - 4}}{\sqrt{1 + m^2}} \right| = 4 \] Simplifying, \[ 495m^4 + 104m^2 - 400 = 0 \] \[ (5m^2 - 4)(99m^2 + 100) = 0 \] \[ m^2 = \frac{4}{5} = \frac{2}{\sqrt{5}} \] Substituting the value of m in (i), we get \[ 2x - \sqrt{5}y + 4 = 0 \] So, the equation of the required common tangent is option B.