(c) Family of circles is

${x^2} + {y^2} – 2x – 4y + 1 + \lambda ({x^2} + {y^2} – 1) = 0$.

$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} – 2x – 4y + (1 – \lambda ) = 0$

${x^2} + {y^2} – \frac{2}{{1 + \lambda }}x – \frac{4}{{1 + \lambda }}y + \frac{{1 – \lambda }}{{1 + \lambda }} = 0$…..(i)

Centre is $\left[ {\frac{1}{{1 + \lambda }},\;\frac{2}{{1 + \lambda }}} \right]$
and radius $ = \sqrt {{{\left( {\frac{1}{{1 + \lambda }}} \right)}^2} + {{\left( {\frac{2}{{1 + \lambda }}} \right)}^2} – \frac{{1 – \lambda }}{{1 + \lambda }}} $

$= \frac{{\sqrt {4 + {\lambda ^2}} }}{{1 + \lambda }}$.

Since it touches the line $x + 2y = 0$, hence

Radius = Perpendicular from centre to the line

$i.e.$, $\left| {\frac{{\frac{1}{{1 + \lambda }} + 2\frac{2}{{1 + \lambda }}}}{{\sqrt {{1^2} + {2^2}} }}} \right| = \frac{{\sqrt {4 + {\lambda ^2}} }}{{1 + \lambda }} $

$\Rightarrow \sqrt 5 = \sqrt {4 + {\lambda ^2}} $

$\Rightarrow \lambda \pm 1$

$ \Rightarrow \sqrt 5 = \sqrt {4 + {\lambda ^2}}$

$\Rightarrow \lambda = \pm 1$

$\lambda = – 1$ cannot be possible in case of circle. So $\lambda = 1$.

Thus, from (i) ${x^2} + {y^2} – x – 2y = 0$ is the required equation of the circle.