Question

Let $T_1, T_2$ and be two tangents drawn from $(-2,0)$ onto the circle $C : x^2 + y^2 = 1.$ Determine the circles touching $C$ and having $T_1, T_2$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time.

Answer 1

From figure it is clear that, $\triangle O L S$ is a right triangle with right angle at $L$.
Also, $O L=1$ and $O S=2$
$\therefore 1 \sin (\angle L S O)=\frac{1}{2} \Rightarrow \angle L S O=30^{\circ}$
Since, $S A_{1}=S A_{2}, \Delta S A_{1} A_{2}$ is an equilateral triangle.
The circle with centre at $C_{1}$ is a circle inscribed in the $\triangle S A_{1} A_{2}$. Therefore, centre $C_{1}$ is centroid of $\Delta S A_{1} A_{2}$. This, $C_{1}$ divides $S M$ in the ratio $2: 1$. Therefore, coordinates of $C_{1}$ are $(-4 / 3,0)$ and its radius $=C_{1} M=1 / 3$
$\therefore$ Its equation is $(x+4 / 3)^{2}+y^{2}=(1 / 3)^{2} \ldots$ (i)
The other circle touches the equilateral triangle $S B_{1} B_{2}$ externally. Its radius $r$ is given by $r=\frac{\Delta}{s-a}$,
where $B_{1} B_{2}=a$. But $\Delta=\frac{1}{2}(a)(S N)=\frac{3}{2} a$
and $ s-a=\frac{3}{2} a-a=\frac{a}{2}$
Thus, $r=3$
$\Rightarrow$ Coordinates of $C_{2}$ are $(4,0)$.
$\therefore$ Equation of circle with centre at $C_{2}$ is
$(x-4)^{2}+y^{2}=3^{2} .....$(ii)
Equations of common tangents to circle (i) and circle $C$ are
$x=-1 $ and $ y=\pm \frac{1}{\sqrt{3}}(x+2) \left[T_{1} \text { and } T_{2}\right]$
Equation of common tangents to circle (ii) and circle $C$ are
$x=1 $ and $ y=\pm \frac{1}{\sqrt{3}}(x+2) \left[T_{1} \text { and } T_{2}\right]$
Two tangents common to (i) and (ii) are $T_{1}$ and $T_{2}$ at $O$. To find the remaining two transverse tangents to (i) and (ii), we find a point I which divides the joint of $C_{1} C_{2}$ in the ratio $r_{1}: r_{2}=1 / 3: 3=1: 9$
Therefore, coordinates of I are $(-4 / 5,0)$
Equation of any line through $I$ is $y=m(x+4 / 5)$. It will touch (i) if
$\frac{\mid m\left(\frac{-4}{3}+\frac{4}{5}\right)}{\sqrt{1+m^{2}}}=\frac{1}{3} \Rightarrow\left|-\frac{8 m}{15}\right|=\frac{1}{3} \sqrt{1+m^{2}}$
$\Rightarrow 64 m^{2}=25\left(1+m^{2}\right)$
$\Rightarrow 39 m^{2}=25$
$\Rightarrow m=\pm \frac{\sqrt{1}}{\sqrt{34}}$
Therefore, these tangents are
$y=\pm \frac{5}{\sqrt{39}}\left(x+\frac{4}{5}\right)$