Question

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersects at the points $A$ and $B.$ Then

• A. equation of common tangents will be $2x\pm \sqrt{5}y+4=0$
• B. point of intersection of common tangents is $(-2,0)$
• C. equation of common tangents will be $2x\pm \sqrt{5}y-4=0$
• D. intersection point of common tangents will be $(2,0)$

Answer 1

Answer: (A), (B)

point of intersection of common tangents is $(-2,0)$

A tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ is
$y=mx\pm \sqrt{9m^2-4}$

It is a tangent to $x^2+y^2-8x=0$ also
$\therefore \frac{4m\pm \sqrt{9m^2-4}}{\sqrt{1+m^2}}=4$
$\Rightarrow 4m\pm \sqrt{9m^2-4}=4\sqrt{1+m^2}$
Squaring both sides, we get
$16m^2\pm 8m\sqrt{9m^2-4}+9m^2-4=16(1+m^2)$
$\Rightarrow 9m^2-20=\pm 8m\sqrt{9m^2-4}$
Again squaring both sieds, we get
$\Rightarrow 495m^4+104m^2-400=0$
$\Rightarrow m^2=\frac{4}{5},-\frac{100}{99}$
$\therefore m=\pm \frac{2}{\sqrt{5}}$
Hence possible diagram will be


$\therefore$ Equations of common tangents are
$y=\frac{2}{\sqrt{5}}x+\frac{4}{\sqrt{5}},y=-\frac{2}{\sqrt{5}}x-\frac{4}{\sqrt{5}}$
$\Rightarrow 2x\pm \sqrt{5}y+4=0$
and intersection point of tangents will be $(-2,0)$