• Centre of the director circle to a given circle is same as the centre of the given circle.
• Radius of the director circle is $\sqrt{2}$ times the radius of the given circle.

$d^2=\frac{\left[m\left(x_0-\alpha\right)-\left(y_0-\beta\right)\right]^2}{\left(1+m^2\right)}$

CASE-I

Circles neither touch nor intersect

$C_1:\left(x_1, y_1\right)$

$C_2: \left(x_2, y_2\right)$

$\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} >r_1+r_2$

$\triangle P B C_2 \sim \triangle P A C_1$

$\frac{P c_1}{P c_2}=\frac{r_1}{r_2}$

$\frac{l_1+l_2}{l_2}=1+\frac{l_1}{l_2}=\frac{r_1}{r_2}$

$l_1 / l_2=\left(r_1-r_2\right) / r_2$

$\Rightarrow \quad l_2=\frac{l_1 r_2}{\left(r_1-r_2\right)} \Rightarrow l_2^2=\frac{l_1^2 r_2^2}{\left(r_1-r_2\right)^2}$

$\frac{\left(\beta-y_1\right)}{\left(\alpha-x_1\right)}=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}$

$\quad\quad\quad\quad =\frac{\left(\beta-y_2\right)}{\left(\alpha-x_2\right)}$

$l_2^2=\left(\beta-y_2\right)^2 +\left(\alpha-x_2\right)^2$

$l_2^2=\left(\beta-y_2\right)^2+\left(\alpha-x_2\right)^2=\left(\alpha-x_2\right)^2 \times{\left[1+\frac{\left(\beta-y_2\right)^2}{\left(\alpha-x_2\right)^2}\right]}$

$\Rightarrow l_2^2=\left(\alpha-x_2\right)^2\left[1+\frac{\left(y_2-y_1\right)^2}{\left(x_2-x_1\right)^2}\right]$

$=\frac{l_1^2 r_2^2}{\left(r_1-r_2\right)^2}=\frac{r_2^2}{\left(r_1-r_2\right)^2}\left[\left(y_2-y_1\right)^2+\left(x_2-x_1\right)^2\right]$

$\frac{(\alpha-x_2 )^2}{ (x_2-x_1 )^2} [ (x_2-x_1 )^2+ (y_2-y_1)^2 ]$$$$=\frac{r_2^2}{ (r_1-r_2 )^2} \times [ (x_2-x_1 )^2+ (y_2-y_1)^2 ]$

$\alpha-x_2=\frac{r_2}{\left(r_1-r_2\right)}\left(x_2-x_1\right)$

$\alpha= x_2+\frac{r_2\left(x_2-x_1\right)}{\left(r_1-r_2\right)}$

$\alpha= \left(r_1 x_2-r_2 x_1\right) /\left(r_1-r_2\right)$

$\beta= \left(r_1 y_2-r_2 y_1\right) /\left(r_1-r_2\right)$

$\frac{(y-\beta)}{(x-\alpha)} = m$

$\frac{\left(\beta -y_1\right)}{\left(\alpha-x_1\right)}=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}$

As Shortest- distance of a point-from a straight line:

$d^2=\frac{\left[m\left(x_0-\alpha\right)-\left(y_0-\beta\right)\right]^2}{\left(1+m^2\right)}$

$\Rightarrow \frac{\left[m\left(x_1-\alpha\right)-\left(y_1-\beta\right)\right]^2}{\left(1+m^2\right)}=r_1^2 \ldots(i)$

$\frac{\left[m\left(x_2-\alpha\right)-\left(y_2-\beta\right)\right]^2}{\left(1+m^2\right)}=r_2^2 \ldots(ii)$

$(x_2-\alpha)^2 \frac{\left[m-\frac{(y-\beta)}{(x_2-\alpha)}\right]^2}{(1+m^2)} =r_2^2$

$\frac{\left[m-\frac{(y_2-\beta)}{(x_2-\alpha)}\right]^2}{[1+m^2]}=\frac{r_2^2}{(x_2-\alpha)^2}\ldots({ii}^{\prime})$

$\frac{\left[m-\frac{ (y_1-\beta )}{ (x_1-\alpha )} \right]^2}{ (1+m^2 )}=\frac{r_1^2}{ (x_1-\alpha )^2} \ldots (i^{\prime})$

$\frac{ \left[m-\frac{(y-\beta)}{ (x_1-\alpha )} \right]^2}{ (1+m^2 )}=\frac{r_2^2}{ (x_2-\alpha)^2} \ldots (ii^{\prime})$

$\frac{r_1}{r_2}=\frac{P C_1}{P C_2} \Rightarrow \frac{r_1^2}{r_2^2}=\frac{P C_1^2}{P C_2^2}$

$\frac{r_1^2}{r_2^2}=\frac{\left(x_1-\alpha\right)^2+\left(y_1-\beta\right)^2}{\left(x_2-\alpha\right)^2+\left(y_2-\beta\right)^2}=\frac{\left(x_1-\alpha\right)^2\left[1+\frac{\left(y_1-\beta\right)^2}{\left(x_1-\alpha\right)^2}\right]}{\left(x_2-\alpha\right)^2\left[1+\frac{\left(y_2-\beta\right)^2}{\left(x_2-\alpha\right)^2}\right]}$

$s$ : slope of line joining $C_1$ and $C_2$.

$\frac{(m-s)^2}{\left(1+m^2\right)}=\frac{r_1^2}{\left(x_1-\alpha\right)^2}, \alpha=\frac{r_1 x_2-r_2 x_1}{\left(r_1-r_2\right)}$

$\therefore x_1-\alpha=x_1+\frac{\left(r_2 x_1-r_1 x_2\right)}{\left(r_1-r_2\right)}$

$\text { ie., } x_1-\alpha=\frac{r_1\left(x_1-x_2\right)}{\left(r_1-r_2\right)}$

$\frac{(m-s)^2}{\left(1+m^2\right)}=\frac{\left(r_1-r_2\right)^2}{\left(x_1-x_2\right)^2}=k$

$m=m_1,m_2$

$m^2-2 m s+s^2=k+k m^2$

$m^2(k-1)+2 ms+\left(k-s^2\right)=0$

$m_1=m_1, m_2 \cdot$

$(y-\beta)=m_1(x-\alpha),$
$\quad(y-\beta)=m_2(x-\alpha)$

$\frac{P C_1}{P C_2}=\frac{r_1}{r_2}$