Auxiliary Circle and Director Circle of an Ellipse

Two circles are naturally associated with every ellipse, and it is worth keeping their definitions and equations clearly distinct:

• The auxiliary circle, drawn with the major axis as diameter: $x^2+y^2=a^2$. It is used purely as a construction device for the eccentric-angle parametrisation (section 7.2.2, part 6) — every point on the ellipse corresponds to a point on this circle sharing the same eccentric angle $\theta$.
• The director circle, the locus of intersection of perpendicular tangents (section 7.2.7): $x^2+y^2=a^2+b^2$. It is always larger than the auxiliary circle (since $b^2>0$), and it is a genuinely different geometric object with a different defining property.

Worked Example 1 — tangent at a point, and at a given eccentric angle.

1. Ellipse $\dfrac{x^2}{8}+\dfrac{y^2}{6}=1$ ($a^2=8,b^2=6$), tangent at $(2,\sqrt3)$: using $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$: $\dfrac{2x}{8}+\dfrac{\sqrt3\,y}{6}=1 \Rightarrow \dfrac{x}{4}+\dfrac{\sqrt3y}{6}=1$, i.e. $3x+2\sqrt3y=12$.
2. Ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ ($a^2=25,b^2=9$), tangent at eccentric angle $\theta=\pi/4$: using $\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1$: $\dfrac{x}{5}\cdot\dfrac{1}{\sqrt2}+\dfrac{y}{3}\cdot\dfrac{1}{\sqrt2}=1$, i.e. $3x+5y=15\sqrt2$. Worked Example 2 — verifying a given line is tangent. Ellipse $4x^2+9y^2=72$, i.e. $\dfrac{x^2}{18}+\dfrac{y^2}{8}=1$ ($a^2=18,b^2=8$); line $2x+3y=12$, i.e. $y=-\dfrac23x+4$, so $m=-\dfrac23,\,c=4$. Check: $c^2=16$; $a^2m^2+b^2=18\left(\dfrac49\right)+8=8+8=16$. Equal — so the line IS tangent to the ellipse. …