Question

The radius of director circle of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ is

• A. $6$
• B. $7$
• C. $\sqrt{7}$
• D. $8$

Answer 1

Answer: (C)

$\sqrt{7}$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,

The equation of Director circle is given by $x^2+y^2=a^2-b^2$

Here the given hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$,

Here $a=4$ and $b=3$

So equation of the director circle will be $x^2+y^2=(4{)}^2-(3{)}^2$

$\to x^2+y^2=7$

Hence the radius of the director circle is $\sqrt{7}$. So correct option is $C$.