The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$ I have
$(1.) \quad x^2 + y^2 - x - 2y = 0 \\
(2.) \quad x + 2y = c$
Solving for $y$ in $(2.)$ gives
$y = (c - x) / 2$
Is there a way to simplify equation $(1.)$?
Because at the end I arrive at 
$c^2 - 2x - c = 0$
and can't proceed. Need to get typical form of square equation $Ax^2 + Bx + C = 0$.
The solution for $c$ is $0$ or $5$.
EDIT 1:
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) $ \quad x^2 + y^2 - x - 2y = 0, \quad x + 2y = c$,  Ans. c = 0 or 5.
The process is to solve for $y$, then to substitute that into second equation.
From that we get A, B and C. Delta being $B^2 - 4AC$ we can get parameter.
So
$y = (c - x) / 2$
$x^2 + y^2 - x -2y = 0$
$x^2 + ((c-x)/2)^2 - x - 2 * ((c-x)/2) = 0$
and i got lost...
 A: If you substitute correctly, you will get the equation $5\,{x}^{2}-2\,c\,x+{c}^{2}-4\,c=0$.
If you solve this, you will get $x = \frac{1}{5}(c \pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, but you don't need this to answer your question.
If the two sets of solutions have the same values, then we must have $c(5-c)=0$, hence $c=0$ or $c=5$.
Addendum: To get the above equation, replace $y$ in $x^2+y^2-x-2y=0$ by $y = \frac{1}{2}(c-x)$ (from the second equation). That is, the equation
\begin{eqnarray}
x^2+\frac{1}{4}(c-x)^2 -x -(c-x) &=& \frac{1}{4}(4x^2+(c-x)^2 -4c) \\
&=& \frac{1}{4}(5 x^2-2 c x +c^2-4c)
\end{eqnarray}
Then the solutions are (ignoring the $\frac{1}{4}$, of course):
\begin{eqnarray}
x &=& \frac{1}{10}\left(2c \pm \sqrt{4 c^2-20(c^2-4c)} \right) \\
&=& \frac{1}{10}\left(2c \pm \sqrt{16 c(5-c)} \right) \\
&=& \frac{1}{5}\left(c \pm 2\sqrt{ c(5-c)} \right) 
\end{eqnarray}
A: You're intersecting a circle with the straight lines parallel to $x+2y=0$.
The circle can be written, by completing the squares, as
$$
x^2 -2\frac{1}{2} x + \frac{1}{4} + y^2 - 2y + 1 = \frac{1}{4}+1
$$
or, writing the squares,
$$
\biggl(x-\frac{1}{2}\biggr)^2+(y-1)^2=\frac{5}{4}
$$
Thus the lines you're looking for must have distance $\sqrt{5}/2$ from the center $(1/2,1)$.
The formula for the distance of a point from a line gives then
$$
\frac{\left|\frac{1}{2}+2\cdot1-c\right|}{\sqrt{1^2+2^2}}=\frac{\sqrt{5}}{2}
$$
that becomes
$$
2\left|\frac{5}{2}-c\right|=5.
$$
This in turn gives the two equations
$$
5-2c=5,\qquad -5+2c=5
$$
that have solutions, respectively, $c=0$ and $c=5$.
You can also solve the system by substitution. From the second equation you get $x=c-2y$ and, substituting in the quadratic equation,
$$
(c-2y)^2+y^2-(c-2y)-2y=0
$$
Developing and simplifying, you get
$$
5y^2-4cy+c^2-c=0
$$
that has coincident roots when the discriminant is zero:
$$
(4c)^2-4\cdot 5(c^2-c)=0
$$
that becomes
$$
-4c^2+20c=0
$$
that is, $c^2-5c=0$. The solutions are $c=0$ and $c=5$.
