Determining the fourth vertex of a parallelogram knowing that its the point of intersection of two circles This question was part of the exercises in one of the courses i'm taking. The answer was already provided. The first circle was assumed to have as its center, vector $v_1$, while its radius was the distance between vectors $v_2$ and $v_3$. The second circle was assumed to have as its center, vector $v_3$, while the distance between vector $v_1$ and $v_2$ was its radius. This gave two equations and we solve the equations to get $D$. What I don't understand is why the distance between the vectors is assumed to be the radius. I have an inkling that this has to do with the properties of the parallelogram but i don't really know. Any input would be appreciated.
Question: Find the fourth vertex $D(x,y)$ of the parallelogram determined by the vertices $A(2,1)$, $B(5,3)$, and $C(7,8)$ as a point of intersection of two circles determined by the vectors:
$$
v_1= [2,1]^T, v_2=[5,3]^T, v_3=[7,8]^T
$$

We see the desired parallelogram in this diagram. (I emphasize the points here rather than the vectors.)

The relevant property of a parallelogram here is that opposites sides have the same length. Therefore, the length of segment $AB$ equals the length of segment $CD$. One way to find point $D$ is to draw the circles with the appropriates radii to guarantee those equal segments, so that means drawing a circle centered at point $A$ with radius $BC$ and a circle centered at $C$ with radius $AB$.
You did not ask for help in solving the two circle equations to find point $D$: ask if you do need that help.
