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, $△OLS$ is a right triangle with right angle at $L$.
Also, $OL=1$ and $OS=2$
$\therefore 1\sin (\angle LSO)=\frac{1}{2}\Rightarrow \angle LSO=30^{\circ }$
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 $△S{A}_1{A}_2$. Therefore, centre $C_1$ is centroid of $\Delta S{A}_1{A}_2$. This, $C_1$ divides $SM$ 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 $(x+4/3{)}^2+y^2=(1/3{)}^2\dots$ (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 $\Delta =\frac{1}{2}(a)(SN)=\frac{3}{2}a$
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
$\frac{\mid m\left(\frac{-4}{3}+\frac{4}{5}\right)}{\sqrt{1+m^2}}=\frac{1}{3}\Rightarrow \left|-\frac{8m}{15}\right|=\frac{1}{3}\sqrt{1+m^2}$
$\Rightarrow 64m^2=25\left(1+m^2\right)$
$\Rightarrow 39m^2=25$
$\Rightarrow m=\pm \frac{\sqrt{1}}{\sqrt{34}}$
Therefore, these tangents are
$y=\pm \frac{5}{\sqrt{39}}\left(x+\frac{4}{5}\right)$