Find all values of $a$ such that this system of equation has solutions
\begin{align}
x^{2}+y\phantom{^2} &=2 x+a \\
x^{2}+y^{2}&=2 x
\end{align}
I tried to substitute second equation into the first one, I managed to make a quadratic equation and I got that $a$ must be less or equal to $1/4$.
but the correct answer is $[-2;1/4]$
I also graphed that and I could see that the minimum value of is actually -2 but how to prove this? What do I miss?
Best Answer
Here is a different approach for the sake of curiosity.
According to the second equation, we have that \begin{align*} x^{2} + y^{2} = 2x & \Longleftrightarrow (x - 1)^{2} + y^{2} = 1 \end{align*}
So the solutions must belong to the corresponding circle.
On the other hand, the first equation reduces to \begin{align*} y = -(x - 1)^{2} + 1 + a \end{align*}
which is the parabola $y = -(x - 1)^{2}$ translated by $1 + a$.
If $1 + a \geq -1$, then the proposed parabola does not touch the circle.
Hence we conclude that $1 + a \geq -1$.
Now it remains to find the other bound for $a$.
In order to find the upper bound for $a$, we are interested in determining for which values of it such that the parabola still touches the circle. This corresponds to find out the the maximum value of $a$ such that the next inequation holds:
\begin{align*} -(x - 1)^{2} + 1 + a \leq \sqrt{1 - (x - 1)^{2}} & \Longleftrightarrow 1 + a \leq \sqrt{1 - (x - 1)^{2}} + (x - 1)^{2} \end{align*}
Once the maximum of the RHS is given by $5/4$, the other bound for $a$ is given by $1 + a \leq 5/4$, that is to say, $a \leq 1/4$.
Hence the desired solution is given by $a\in[-2,1/4]$.