Director Circle

Locus of points from which the two tangents to x²/a² − y²/b² = 1 are perpendicular. Real only when a > b. When a = b (rectangular hyperbola), the director circle degenerates to a point (the centre). When a < b, no real director circle exists.

Let $P(h, k)$ be a point from which the two tangents to $x^2/a^2 - y^2/b^2 = 1$ are perpendicular.

The tangent $y = mx + \sqrt{a^2m^2-b^2}$ passes through $(h,k)$:

$$(k-mh)^2 = a^2m^2-b^2$$ $$m^2(h^2-a^2) - 2mhk + (k^2+b^2) = 0$$

This quadratic in $m$ gives slopes $m_1, m_2$ with:

$$m_1m_2 = \frac{k^2+b^2}{h^2-a^2}$$

For perpendicular tangents: $m_1m_2 = -1$:

$$\frac{k^2+b^2}{h^2-a^2} = -1 \implies k^2+b^2 = -(h^2-a^2) = a^2-h^2$$ $$h^2+k^2 = a^2-b^2$$

Locus: $x^2+y^2 = a^2-b^2$.

Three cases:

• $a > b$: real director circle, radius $\sqrt{a^2-b^2}$ (smaller than the auxiliary circle)
• $a = b$ (rectangular hyperbola): $x^2+y^2 = 0$, degenerates to the origin — only point from which perpendicular tangents can be drawn is the centre
• $a < b$: $a^2-b^2 < 0$ — no real locus, no pair of perpendicular tangents exists from any real point