Director Circle

Locus of points from which the two tangents to the ellipse are perpendicular. A concentric circle of radius √(a²+b²). For a circle (a = b), the director circle has radius a√2.

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

Any tangent with slope $m$ is $y = mx + \sqrt{a^2m^2+b^2}$. For this to pass through $P(h,k)$:

$$k = mh + \sqrt{a^2m^2+b^2}$$ $$(k - mh)^2 = a^2m^2 + b^2$$ $$k^2 - 2mhk + m^2h^2 = a^2m^2 + b^2$$ $$m^2(h^2-a^2) - 2hkm + (k^2-b^2) = 0$$

This quadratic in $m$ gives slopes $m_1$ and $m_2$ of the two tangents from $P$.

By Vieta's formulas:

$$m_1 + m_2 = \frac{2hk}{h^2-a^2}, \quad m_1m_2 = \frac{k^2-b^2}{h^2-a^2}$$

For perpendicular tangents: $m_1m_2 = -1$:

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

The locus (replacing $h \to x$, $k \to y$) is:

$$x^2 + y^2 = a^2 + b^2$$

Note: For a circle with $a = b = r$, this gives $x^2+y^2 = 2r^2$ — consistent with the director circle of a circle.