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Question: The radius of the director circle of hyperbola \[\dfrac{\mathrm x^2}{\mathrm a^2}-\dfrac{\mathrm y^2...

The radius of the director circle of hyperbola x2a2−y2b2=1\dfrac{\mathrm x^2}{\mathrm a^2}-\dfrac{\mathrm y^2}{\mathrm b^2}=1 is
A. a
B. b
C.  a2+b2D.  a2−b2\mathrm C.\;\sqrt{\mathrm a^2+\mathrm b^2}\\\\\mathrm D.\;\sqrt{\mathrm a^2-\mathrm b^2}

Explanation

Solution

Hint:The director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. The equation of the director circle of a general hyperbola is given by-x2+y2=a2−b2x^2 + y^2 = a^2 - b^2.Comparing it with the general equation of circle we get the radius of the director circle of hyperbola.

Complete step-by-step answer:
The given equation of hyperbola is a general equation. So the equation of its director circle is given by-
x2+y2=a2−b2x^2 + y^2 = a^2 - b^2.....(1)

Now, we have to find the radius of this circle. The general equation of a circle at centre (0, 0) is given by x2+y2=r2x^2 + y^2 = r^2
, where r is the radius.
By comparing (1) and (2),
r2=a2−b2r^2 = a^2 - b^2
r=a2−b2\mathrm r=\sqrt{\mathrm a^2-\mathrm b^2}
Hence the Director circle is a circle whose centre is same as centre of the hyperbola and the radius is a2−b2\sqrt{\mathrm a^2-\mathrm b^2}
This is the required answer, the correct option is D.

Note: One should know the definition and formula of the director circle.Even if formula is not known, we can use the definition to find the locus of points that form a director circle.The locus of point of intersections of perpendicular tangents to the hyperbola is a circle called as Director circle and for a standard hyperbola x2a2−y2b2=1\dfrac{\mathrm x^2}{\mathrm a^2}-\dfrac{\mathrm y^2}{\mathrm b^2}=1, its equation is x2+y2=a2−b2x^2+y^2 = a^2-b^2.