Find min(P).

Here is a common incorrect solution that solvers should avoid:

$\begin{cases} x+y=2a-1 \\ { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }+2a-3 \end{cases}\\ \Rightarrow (x+y)^{ 2 }-({ x }^{ 2 }+{ y }^{ 2 })={ (2a-1) }^{ 2 }-({ a }^{ 2 }+2a-3)\\ \Leftrightarrow 2xy=(4{ a }^{ 2 }-4a+1)-({ a }^{ 2 }+2a-3)\\ \Leftrightarrow 2P=3{ a }^{ 2 }-6a+4\\ \Leftrightarrow 2P=3(a-1)^{ 2 }+1\ge 1\\ \Rightarrow P\ge 0.5$

However, $P=0.5$ occurs when ${ (a-1) }^{ 2 }=0\Leftrightarrow a=1$ . In this case, $\begin{cases} x+y=1 \\ { x }^{ 2 }+{ y }^{ 2 }=0 \end{cases}$ , which is impossible.

Here is the correct solution :

As proven in the solution above, $2P=3{ a }^{ 2 }-6a+4\\ \Rightarrow 4P=6{ a }^{ 2 }-12a+8$

Suppose $S=x+y$ . The given system of equations has at least a solution when $S^{ 2 }-4P\ge 0$ (According to the inverse of Vieta Formula)

$\Leftrightarrow (2a-1)^{ 2 }-(6{ a }^{ 2 }-12a+8)\ge 0\\ \Leftrightarrow -2{ a }^{ 2 }+8a-7\ge 0\\ \Leftrightarrow -2({ a }^{ 2 }-4a+4)\ge -1\\ \Leftrightarrow { (a-2) }^{ 2 }\le \frac { 1 }{ 2 } \\ \Leftrightarrow -\frac { \sqrt { 2 } }{ 2 } \le a-2\le \frac { \sqrt { 2 } }{ 2 } \\ \Leftrightarrow \frac { 2-\sqrt { 2 } }{ 2 } \le a-1\le \frac { 2+\sqrt { 2 } }{ 2 } \\ \Rightarrow 3\times (\frac { 2-\sqrt { 2 } }{ 2 } )^{ 2 }+1\le 3{ (a-1) }^{ 2 }+1\\ \Rightarrow P\ge \frac { 11-6\sqrt { 2 } }{ 4 } \\$

Hence, the solution is $\frac { 11-6\sqrt { 2 } }{ 4 }$

By substitution, $x^2 + (2a - 1 - x)^2 = a^2 + 2a - 3$ , which simplifies to $2x^2 + (2 - 4a)x + 3a^2 - 6a + 4 = 0$ , which solves to $x = \frac{2a - 1 \pm \sqrt{-2a^2 + 8a - 7}}{2}$ . Therefore, $x$ has real solutions when $-2a^2 + 8a - 7 > 0$ , or when $\frac{4 - \sqrt{2}}{2} \leq a \leq \frac{4 - \sqrt{2}}{2}$ .

Now $P = xy = \frac{1}{2}((x + y)^2 - (x^2 + y^2))$ which by substitution is $P = \frac{1}{2}((2a - 1)^2 - (a^2 + 2a - 3))$ or $P = \frac{3}{2}(a - 1)^2 + \frac{1}{2}$ , which is increasing for $\frac{4 - \sqrt{2}}{2} \leq a \leq \frac{4 - \sqrt{2}}{2}$ . Therefore, the minimum value of $P$ is when $a = \frac{4 - \sqrt{2}}{2}$ at $P = \frac{3}{2}(\frac{4 - \sqrt{2}}{2} - 1)^2 + \frac{1}{2} = \frac{11 - 6\sqrt{2}}{4} \approx \boxed{0.629}$ .