13. As shown in the figure, the line $x = 2$ intersects the asymptotes of the hyperbola $\Gamma : \frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$ at points $E _ { 1 }$ and $E _ { 2 }$. Let $\overrightarrow { O E _ { 1 } } = \overrightarrow { e _ { 1 } }$ and $\overrightarrow { O E _ { 2 } } = \overrightarrow { e _ { 2 } }$. For any point $P$ on the hyperbola $\Gamma$, if $\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } }$ ($a , b \in \mathbb{R}$), then $a$ and $b$ satisfy the equation $\_\_\_\_$ $4 a b = 1$. Analysis: $E _ { 1 } ( 2,1 )$, $E _ { 2 } ( 2 , - 1 )$ $$\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } } = ( 2 a + 2 b , a - b ), \text{ point } P \text{ is on the hyperbola}$$ $\therefore \frac { ( 2 a + 2 b ) ^ { 2 } } { 4 } - ( a - b ) ^ { 2 } = 1$, which simplifies to $4 a b = 1$ [Figure]
(5 points) The equation of the circle with center at the origin that is tangent to the line $x + y - 2 = 0$ is $\_\_\_\_$.
13. As shown in the figure, the line $x = 2$ intersects the asymptotes of the hyperbola $\Gamma : \frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$ at points $E _ { 1 }$ and $E _ { 2 }$. Let $\overrightarrow { O E _ { 1 } } = \overrightarrow { e _ { 1 } }$ and $\overrightarrow { O E _ { 2 } } = \overrightarrow { e _ { 2 } }$. For any point $P$ on the hyperbola $\Gamma$, if $\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } }$ ($a , b \in \mathbb{R}$), then $a$ and $b$ satisfy the equation $\_\_\_\_$ $4 a b = 1$.
Analysis: $E _ { 1 } ( 2,1 )$, $E _ { 2 } ( 2 , - 1 )$
$$\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } } = ( 2 a + 2 b , a - b ), \text{ point } P \text{ is on the hyperbola}$$
$\therefore \frac { ( 2 a + 2 b ) ^ { 2 } } { 4 } - ( a - b ) ^ { 2 } = 1$, which simplifies to $4 a b = 1$\\
\includegraphics[max width=\textwidth, alt={}, center]{92627018-2332-482e-855e-1d26e13a1d2f-3_369_462_1731_1407}