Explanation

(b)
Equation of any circle through (0, 0) and (1, 0) is
$(x - 0)(x - 1) + (y - 0)(y - 0) + \lambda\left| \begin{matrix}
x & y & 1 \\
0 & 0 & 1 \\
1 & 0 & 1
\end{matrix} \right| = 0$
$\Rightarrow x^{2} + y^{2} - x + \lambda y = 0$
If it represents $C_{3}$, its radius=1
$\Rightarrow 1 = \left( \frac{1}{4} \right) + \left( \frac{\lambda^{2}}{4} \right)\ \Rightarrow \ \lambda = \pm \sqrt{3}$
[image:11_image23.emf]
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 $\left( \frac{1}{2},\frac{\sqrt{3}}{1} \right)$
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, we observe that the required tangent makes positive intercept on the $y$-axis and negative on the $x$-axis and hence, its equation is $\sqrt{3}x - y + 2 = 0$