Calculate the middle point of the segment joining both centers: $M=\left(\frac{10+5}{2}\,,\,\frac{10+3}{2}\right)=\left(\frac{15}{2}\,,\,\frac{13}{2}\right)$
The point $\,M\,$ is the point of intersection of the rectangle-to-be's diagonals (can you see why?).
Now the distance between two centers: $d=\sqrt{(10-5)^2+(3-10)^2}=\sqrt{74}$ so the rectangle's long side is of length $\,\sqrt {74}+10\,$ and its width is, of course, $\,10\,$.
Can you take it from here? For example, both long sides of the rectangle are, of course, parallel but also parallel with the segment joining the circles' lengths. This last has slope equal to $m=\frac{10-3}{5-10}=-\frac{7}{5}...$
Added as answer to the OP: Point of origin?? You only need to find the equation of the line through, say the upper circle's center, and which is perpendicular to both long sides of the rectangle and then find both points on it at a distance of $\,5\,$ from the circle's center. These two points are the intersection points of the rectangle's long sides with that circle, and now you can easily find the red point on the rectangle by finding the points 5 units away from these intersection points and on the upper long side of the rectangle..
Another way: Draw the triangle between both upper red points and the intersection point of the circle with the upper long side of the rectangle. You get an isosceles straight-angle triangle with legs with lengths 5 (why?), so the red point on the rectangle is on the line forming an angle of $\,45^\circ\,$ with the long side of the rectangle (or with the radius you found above) and at distance of $\,5\sqrt 2\,$ (Pythagoras is our friend here) from the circle's center...I hope you're doing the maths and the drawings as you read this since it is waaaaaaayy harder to write it down than to understand it: it aall is very basic geometry + basic analytic geometry.