Director Circle of Hyperbola

In Geometry, we have learned different types of conic sections and hyperbola is one among them. A hyperbola is defined as the locus of points whose distance from the fixed point called focus from the fixed-line called directrix is a constant. For hyperbola, the eccentricity is greater than 1. (i.e) e>1. In this article, we are going to discuss one of the parameters called the “Director Circle of Hyperbola” in detail with many solved examples.

What is the Director Circle of Hyperbola?

The director circle of the hyperbola is defined as a locus of the point of intersection of the two perpendicular tangents to the hyperbola. We know that the standard equation of hyperbola is (x²/a²) – (y²/b²) = 1. Thus, the equation of the director circle of a hyperbola is derived from the standard form.

Equation of Director Circle of Hyperbola

The equation of the director circle of the hyperbola (x²/a²) – (y²/b²) = 1 is x²+y² = a² – b².

Where, 

The centre of the circle is the origin (i.e.) (0, 0) and the radius is √(a²-b²)

Three Cases of Director Circle of Hyperbola

Case 1: b² = a²

If b² = a², then the radius of the circle is zero, and it reduces the point at the centre of the circle (origin). So, in this case, the centre is the only point from where we can draw a tangent at the right angle to the hyperbola.

Case 2: b² < a²

If b² < a², then the director circle of the hyperbola is real.

Case 3: b²>a²

If b² > a², the radius of the circle is imaginary. In this case, there should not be any circle and no tangents at right angles can be drawn to the circles.

Director Circle of Hyperbola Examples

Go through the following examples to understand the director circle of a hyperbola:

Example 1:

Determine the equation of the director circle of the hyperbola (x²/16) – (y²/4) = 1

Solution:

Given: Hyperbola equation is (x²/16) – (y²/4) = 1

We know that the equation of the director circle of the hyperbola (x²/a²) – (y²/b²) = 1 is x²+y² = a² – b².

 By comparing the given equation and standard form, we have a² = 16 and b² = 4

⇒ x²+y² = 16 – 4

⇒ x²+y² = 12.

Therefore, the required equation for the director circle of a hyperbola is x²+y² = 12.

Example 2:

Find the equation and the diameter of the director circle of the hyperbola (x²/49) – (y²/25) = 1

Solution:

Given hyperbola equation: (x²/49) – (y²/25) = 1 …(1)

The standard form of hyperbola is (x²/a²) – (y²/b²) =1  …(2)

As we know, the director cirlce of hyperbola equation is x²+y² = a²-b²

By comparing (1) and (2), we get

a² = 49 and b² = 25

Thus, the required director circle of hyperbola equation is x²+y² = 49 – 25 

⇒ x²+y² = 24.

We know that the radius of the director circle = √(a²-b²) = √24.

As the diameter is twice the radius,  the diameter of the director circle of the hyperbola is 2√24.

Example 3:

Find the equation of director circle of the hyperbola (x²/100) – (y²/36) = 1

Solution:

Given: (x²/100) – (y²/36) = 1

Here, a² = 100 and b² = 36

Substituting the values in the director cirlce equation, we get

⇒ x²+y² = 100 – 36

⇒ x²+y² = 64, which is the required director circle equation of hyperbola.

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