Director Circle of Ellipse

An ellipse is the locus of a point which moves such that its distance from a fixed point (focus) gives a constant. The tangent of an ellipse is a line which touches the ellipse at only one point without passing through it. This concept of tangent is used in director circles. we use the director circle to determine important properties of the ellipse. In this article, we discuss the director circle of ellipses which falls under the topic of dimensional Analytical Geometry.

What is the Director Circle of an Ellipse?
The director circle of the ellipse is the locus of the point of intersection of the perpendicular tangents of the ellipse at right angles.

Equation of the Director Circle of an Ellipse

The equation of the director circle of the ellipse $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with centre as origin $(0,0)$ is $x^2+y^2=a^2+b^2$.
When the center of the ellipse is not at the origin but at $(h, k)$, then the equation becomes $(x-h)^2+(y-k)^2=a^2+b^2$
where a and b are the lengths of the semi-major and semi-minor axes, respectively.

Derivation of Equation of Director Circle of an Ellipse

Equation of tangent of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in slope form is $y=m x+\sqrt{a^2 m^2+b^2}$ passing through the point $(h, k)$

$
\begin{aligned}
& k=m h+\sqrt{a^2 m^2+b^2} \\
& (k-m h)^2=a^2 m^2+b^2 \\
& k^2+m^2 h^2-2 m h k=a^2 m^2+b^2 \\
& \left(h^2-a^2\right) m^2-2 h k m+k^2-b^2=0
\end{aligned}
$

This is quadratic equation in $m$, slope of two tangents are $m_1$ and $m_2$

$
\begin{aligned}
& \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{k}^2-\mathrm{b}^2}{\mathrm{~h}^2-\mathrm{a}^2} \\
& -1=\frac{\mathrm{k}^2-\mathrm{b}^2}{\mathrm{~h}^2-\mathrm{a}^2} \quad[\text { tangents are perpendicular }] \\
& -\mathrm{h}^2+\mathrm{a}^2=\mathrm{k}^2-\mathrm{b}^2 \\
& \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2+\mathrm{b}^2
\end{aligned}
$

An ellipse having a fixed length of major and minor axes slides between two perpendicular straight lines. So, these lines are perpendicular tangents and their point of intersection P lies on the director circle. If the centre of the ellipse is fixed, then all the points of intersection of perpendicular tangents lie at a fixed distance which is equal to the radius of the director circle. So, if the point of intersection of perpendicular tangents is fixed (point $P$ ), then the centre of the variable ellipse also lies at a fixed distance from point $P$. Therefore, the locus of the centre of the ellipse is a circle

Solved Examples

Example 1: The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Find the equation of the director circle for the ellipse.

Solution:
For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the equation of the director circle is:

$
x^2+y^2=a^2+b^2
$
Here, $a^2=25$ and $b^2=16$.

Radius,

$
a^2+b^2=25+16=41
$
Substituting the value,

The director circle of ellipse

$
x^2+y^2=41
$

Example 2: Find the points intersecting the coordinate axes and the director circle of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$

Solution:
The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$,

Then, $a^2=9$ and $b^2=4$.

The equatio of the director circle,

$
x^2+y^2=a^2+b^2=9+4=13
$
When it on the x -axis, $(y=0)$

$
x^2=13 \quad \Rightarrow \quad x= \pm \sqrt{13}
$

When intersecting on the $y$-axis $(x=0)$ :

$
y^2=13 \quad \Rightarrow \quad y= \pm \sqrt{13}
$

Hence, the points of intersection are $( \pm \sqrt{13}, 0)$ and $(0, \pm \sqrt{13})$.

Example 3: Find the equation of the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{25}=1$ that touches the director circle.

Solution:
The given equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{25}=1$

The equation of the director circle of the give ellipse is

$
x^2+y^2=36+25=61
$

A tangent to the ellipse can be written as:

$
\frac{x x_1}{36}+\frac{y y_1}{25}=1
$

When the tangent touches the director circle, the radius of the director circle is the distance between the cneter of the ellipse and the point of contact.

The distance from the center of the ellipse $(0,0)$ to the tangent line $a x+b y=c$ is:

$
\frac{|c|}{\sqrt{a^2+b^2}}=\sqrt{61}
$

$
\frac{c}{\sqrt{a^2+b^2}}=\sqrt{61}
$

The equation of the tangent will be:

$
\frac{x x_1}{36}+\frac{y y_1}{25}=\sqrt{61}
$

Example 4: Find the area of the director circle of the ellipse $\frac{x^2}{5}+\frac{y^2}{4}=1$

Solution

The equaiton of ellipse, $\frac{x^2}{5}+\frac{y^2}{4}=1$
The equation of director circle of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2+b^2$

Here, $a^2=5, b^2=4$
Equation of director circle is

$
\begin{aligned}
& x^2+y^2=5+4 \\
& \Rightarrow x^2+y^2=9=r^2 \\
& \therefore \text { Area }=\pi r^2=9 \pi \text { sq. units }
\end{aligned}
$

Summary
The director circle of the ellipse is the locus of the point of intersection of the perpendicular tangents of the ellipse of right angles. The equation of the director circle of the ellipse with centre $(h,k)$ is $(x-h)^2+(y-k)^2=a^2+b^2$.