The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$

Question

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...

Answer 1

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}

Answer 2

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 $c=5$.