Director Circle - Practice Questions & MCQ

Director Circle

The locus of the point through which perpendicular tangents are drawn to a given circle $S=0$ is a circle called the director circle of the circle $\mathrm{S}=0$.

If two tangents are drawn from any point on the Director circle to the circle, then the angle between the tangents is $90^{\circ}$.

The equation of director circle of the circle (S) : $x^2+y^2=a^2$ is $\mathrm{x}^2+\mathrm{y}^2=2 \mathrm{a}^2$.

Proof:
The equation of any tangent to the circle $x^2+y^2=a^2$ is

$
y=m x+a \sqrt{\left(1+m^2\right)} \quad[\text { slope form }]
$
Point $\mathrm{P}(\mathrm{h}, \mathrm{k})$ is point of intersection of two tangents, then point P lies on the Eq (i)

$
\begin{aligned}
& k=m h+a \sqrt{\left(1+m^2\right)} \\
& \text { or } \quad(k-m h)^2=a^2\left(1+m^2\right) \\
& \text { or } \quad m^2\left(h^2-a^2\right)-2 m k h+k^2-a^2=0
\end{aligned}
$
This is quadratic equation in $m$, let two roots are $m_1$ and $m_2$
But tangents are perpendiculars, therefore $m_1 m_2=-1$

$
\Rightarrow \frac{k^2-a^2}{h^2-a^2}=-1 \Rightarrow k^2-a^2=-h^2+a^2 \Rightarrow                          h^2+k^2=2 a^2
$
Hence, locus of $P(h, k)$ is $x^2+y^2=2 a^2$

So the director circle for any circle is a circle which is concentric with the given circle and whose radius is $\sqrt{2}$ times the radius of the given circle. This fact is applicable for the circles that have non-origin centres as well.