In a plain with orthogonal coordinate $XOY$, set point $A(a,a)$, and $P$ is a point in function $y=\frac{1}{x}$,where $x>0$. If the distance between $P$ and $A$ is $2\sqrt{2}$.Find all $a$ satisfies this condition. ($a \in \Bbb{R}$).
I can use distance formula between two point to solve this exercise. $$|PA| = \sqrt{\left(x-a\right)^2+\left(\frac{1}{x}-a\right)^2}=\sqrt{x^2+\frac{1}{x^2}-2a\left(x+\frac{1}{x}\right)+2a^2}$$ then let $u = x+ \frac{1}{x}$. we have $u \geq 2$. then can be find out that $a = -1$ or $a = \sqrt{10}$ directly.
Another approach:
Because we want to find $a$, such that the minimum distant($=2\sqrt{2}$) between $A$ and a point $P$ in function $y=1/x$. So If we set $A$ as the center of a circle ,and radius is $2\sqrt{2}$. this circle is tangent to the function $y=\frac{1}{x}$ at least a point , say $(t, \frac{1}{t})$. then we can get two equations:
$$\begin{cases}\sqrt{(a-t)^2+(a-\frac{1}{t})^2}=2\sqrt{2} & (radius)\\a-t-\frac{1}{t^2}(a-\frac{1}{t})=0 & (tangent)\end{cases} $$
but solve this equation group (use mathematica) , It is not the same as the first way. what's wrong with me? Please help me . thanks very much.