Draw a graph for the inequality $(x_1 - \frac{1}{2})^2 + (y_1 - \frac{1}{2})^2 < (x_2 - \frac{1}{2})^2 + (y_2 - \frac{1}{2})^2$

Question

I need to draw a graph for the inequality $(x_1 - \frac{1}{2})^2 + (y_1 - \frac{1}{2})^2 < (x_2 - \frac{1}{2})^2 + (y_2 - \frac{1}{2})^2$, where $x_1 \in [0, \frac{1}{2}]$, $x_2 \in [\frac{1}{2}, 1]$ and $y_1,y_2 \in [0,1]$. After trying, I can't figure out how to draw this. I am not sure whether this could be two circles with the center at $[\frac{1}{2}; \frac{1}{2}]$. Any help or hint would be appreciated.

Answer 1

The left side is the distance between $P_1=(x_1,y_1)$ and $(1/2,1/2)$ squared. The right side is the distance between $P_2=(x_2,y_2)$ and $(1/2,1/2)$ squared.

So you need to illustrate that $P_2$ is farther than $P_1$ from $(1/2,1/2)$. There is no standard way to do this with a plot of the Cartesian plane. But you can imagine some alternative ways to plot this. If $r_1$ and $r_2$ are the distances from $(1/2,1/2)$, then you could plot $r_1<r_2$ in the $r_1r_2$-plane.

You could also make a sequence of $xy$-plane plots where each has some circle around $(1/2,1/2)$ marked as "possibilities for $P_2$" and its interior is shaded and marked "possibilities for $P_1$". The sequence of plots would have expanding circles for $P_2$.

Answer 2

The mathematical expressions on both sides of inequality represent circles with center at the point $(\frac{1}{2}, \frac{1}{2})$. Assume that both are drawn on the same $xy$-plane, just the variables are renamed or relabeled. Then for the given ranges of $x_1$ and $y_1$, the left hand side draws a semi-circle (left part cut vertically). Similarly for the given ranges of $x_2$ and $y_2$, the right hand side draws a semi-circle (right part cut vertically). Finally both together forms a circle centered at the point $(\frac{1}{2}, \frac{1}{2})$ and the radius $r = \frac{1}{2}$.

Answer 3

Considering $\mathbb{R}^2$ we have

$$ C_1\to (x_1-\frac 12)^2+(y_1-\frac 12)^2= r_1^2\\ C_2\to (x_2-\frac 12)^2+(y_2-\frac 12)^2= r_2^2 $$

and then

$$ r_1^2 < r_2^2 \;\; \forall \{r_1, r_2\} $$

so it is all $\left(\mathbb{R}^2-\{\frac 12,\frac 12\}\right)$