Question

$C_1$ and $C_2$ are circles of unit radius with centres at $(0,0)$ and $(1,0)$ respectively. $C_3$ is a circle of unit radius, passes through the centres of the circles $C_1$ and $C_2$ and have its centre above x-axis. Equation of the common tangent to $C_1$ and $C_3$ which does not pass through $C_2$ is

• A. $x-\sqrt{3}y+2=0$
• B. $\sqrt{3}x-y+2=0$
• C. $\sqrt{3}x-y-2=0$
• D. $x+\sqrt{3}y+2=0$

Answer 1

Answer: (B)

$\sqrt{3}x-y+2=0$

Equation of any circle through $(0,0)$ and $(1,0)$ is
$(x-0)(x-l)+(y-0)(y-0)+\lambda \left|\begin{array}{ccc}x& y& 1\\ 0& 0& 1\\ 1& 0& 1\end{array}\right|=0$
$\Rightarrow x^2+y^2-x+\lambda y=0$ If it represents $C_3$ its radius$=1$
$\Rightarrow 1=\left(\frac{1}{4}\right)+\left(\frac{{\lambda }^2}{4}\right)\Rightarrow \lambda =\pm \sqrt{3}$
As the centre of $C_3$ lies above the x-axis, we take $\lambda .=-\sqrt{3}$ and thus an equation of $C_3$ is $x^2+y^2-x-..\sqrt{3}y=0$ Since $C_1$ and $C_3$ intersect and are of unit radius, their common tangents are parallel to the line joining their centres $(0,0)$ and $(\frac{1}{2},\frac{\sqrt{3}}{2})$.
So, let the equation of a common tangent be
$\sqrt{3}x-y+k=0$ It will touch $C_1$ if
$\left|\frac{k}{\sqrt{3+1}}\right|=1\Rightarrow k=\pm 2$
From the figure $\left(16.25\right)$ we observe that the required tangent makes positive intercept on the y-axis and negative on the x-axis and hence its equation is
$\sqrt{3x}-y+2=0$
Ans: B