Question

The circle $x^{2}+y^{2}-8 x=0$ and hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ 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

• A. $2 x-\sqrt{5} y-20=0$
• B. $2 x-\sqrt{5} y+4=0$
• C. $3 x-4 y+8=0$
• D. $4 x-3 y+4=0$

Answer 1

Answer: (B)

A tangent to $\frac{ x ^{2}}{9}-\frac{ y ^{2}}{4}=1$ is
$y = mx +\sqrt{9 m ^{2}-4}, m >0$
It is tangent to $x^{2}+y^{2}-8 x=0$
$\therefore \frac{4 m+\sqrt{9 m^{2}-4}}{\sqrt{1+m^{2}}}=4$
$\Rightarrow 495\, m ^{4}+104 \,m ^{2}-400=0$
$\Rightarrow m ^{2}=\frac{4}{5}$ or $m =\frac{2}{\sqrt{5}}$
$\therefore $ the tangent is $y =\frac{2}{\sqrt{5}} m +\frac{4}{\sqrt{5}}$
$\Rightarrow 2 x -\sqrt{5} y +4=0$