Question

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

Answer 1

Answer: (B)

Equation of tangent to the hyperbola
$\frac{x^{2}}{9}- \frac{y^{2}}{4}=1$ is
$y=mx+\sqrt{9m^{2}-4} \ldots\left(i\right)$
Since, Eq. $\left(i\right)$ tagent to the circle
$\therefore $ Perpendicular distance from centre $\left(0,4,0\right)$ to the circle is equal to the radius of the circle
$\therefore \frac{\left|4m +0+\sqrt{9m^{2}-4}\right|}{\sqrt{1^{2}+m^{2}}}=4$
$\Rightarrow \left(4m+\sqrt{9 m^{2}-4}\right)=4 \sqrt{1+m^{2}}$
$\Rightarrow 16m^{2}+9m^{2}-4+8m \sqrt{9m^{2}-4} $
$=16\left(1+m^{2}\right)$
$\Rightarrow 9m^{2}-20$
$=-8m \sqrt{9m^{2}-4}$
$\Rightarrow 81m^{4}+400-360 m^{2}$
$=64m^{2}\left(9m^{2}-4\right)$
$\Rightarrow 495m^{4}+104m^{2}-400=0$
$\Rightarrow m=2 \sqrt{5}$
$\therefore $ From Eq. $\left(i\right)$, $y=\frac{2x}{\sqrt{3}}+\sqrt{9\times\frac{4}{5}-4} $
$\sqrt{5}y=2\times T+\sqrt{36-20}$
$\Rightarrow 2x-\sqrt{5}y+4=0 $