Question: I want to find the $$\lim\_{x\rightarrow \infty} (\sqrt x-x)$$

What I was thinking was to multiply by the conjugate since that's normally what you do with square roots, So I would get 

$$\lim\_{x\rightarrow\infty} (\sqrt{x}-x)\cdot\frac{(\sqrt{x}+x)}{(\sqrt{x}+x)}$$

$=\lim\_{x\rightarrow\infty} \frac{x-x^2}{\sqrt{x}+x} = \lim\_{x\rightarrow\infty}\frac{x(1-x)}{\sqrt{x} (1+x)} = \lim\_{x\rightarrow\infty} \frac{-x}{\sqrt{x}}$

I guess as x goes to infinity, $\sqrt \infty$ is smaller than $-\infty$, so it blows up to $-\infty$ but I'm unsure if this logic is correct. 

One solution I found was $\lim\_{x\rightarrow\infty} \sqrt x (1-\sqrt x) = (\infty) (1-\infty) = (\infty)(-\infty) = -\infty$

However I'm just uncomfortable actually plugging in $\infty$ for limit values and I also wouldn't think to do this. I would think to multiply by the conjugate

Answer: 
You are correct with your idea stated in the comment, i.e., the line bisects the chord of $(0,0)$ to $(a,b)$, except you didn't mention the line is perpendicular to this chord. This is because the equal length (i.e., the radius) lines from the center of the circle to each end point of this chord, plus the chord line itself, form an isosceles triangle. Thus, the perpendicular bisector of the chord goes through the opposite triangle end-point, i.e., the center of the circle.
Adding to John Omielan's answer, [according to the perpendicular bisector theorem](https://math.stackexchange.com/questions/2295360/prove-the-perpendicular-bisector-of-chord-passes-through-the-centre-of-the-circle), if a point is on the perpendicular bisector, then it is equidistant from the segment's endpoints.

The converse of this theorem shows that if the point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector. Since $(x,y)$ is the centre of the circle passing through $(0,0)$ and $(a,b)$, then the centre also must lie on the perpendicular bisector.