Question

A variable line has intercepts $e$ and $e^{\prime }$ on the coordinate axes, where $\frac{e}{2}$ and $\frac{e^{\prime }}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively. The value of $r$ for which the line always touches the circle $x^2+y^2=r^2$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Since $\frac{e}{2}$ and $\frac{e^{\prime }}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola, therefore $\frac{4}{e^2}+\frac{4}{e^{\prime 2}}=1$
$\Rightarrow \frac{1}{e^2}+\frac{1}{e^{\prime 2}}=\frac{1}{4}$

The equation of the line is
$\frac{x}{e}+\frac{y}{e^{\prime }}=1$
It is tangent to the circle $x^2+y^2=r^2$
So, the distance from the centre $(0,0)$ to the line is equal to the radius $r$
$r=\frac{|\frac{0}{e}+\frac{0}{e^{\prime }}-1|}{\sqrt{{\left(\frac{1}{e}\right)}^2+{\left(\frac{1}{e^{\prime }}\right)}^2}}\Rightarrow r=2$

Hence, the radius of the circle is $r=2$.