geometry
Is there a formula/calculation to work out how to find the other $2$ coordinates for a rectangle if you only have the bottom left and top right coordinates?
e.g. My bottom-left coordinate would be $(-1, 0)$ and top right would be $(3, -2)$, I can work out the midpoint but I can't seem to get my head around how to work out the other $2$ coordinates?
Use the fact that the centre $(h,k)$ of a rectangle is the midpoint of the diagonals:
$h=\frac{0+5}2, k=\frac{5+0}2 $
Alternatively, use the fact the centre $(h,k)$ of a rectangle is equidistant from the corners, so that $$(h-0)^2+(k-0)^2=(h-0)^2+(k-5)^2=(h-5)^2+(k-5)^2$$
Use the 1st two to get $h$ and the last two to get $k.$
In addition to the answer pointed out by @n_b I would like to show how to reach there:
$$\LARGE\boxed{R=\frac H2+\frac{W^2}{8H}}$$
float cos_t = (R-H)/R;
P_x=(rect.width/2);
P_y=-R.cos_t;
Without using calculus(actually it would never be solved with calculus) and only using trigonometry:
The component of R along $\theta$ is $R.\cos\theta$ and perpendicular to it is $R.\sin\theta$
Now remaining component is $R-Rcos\theta=R(1-\cos\theta)$
Now we have: $$ W=2.R.\sin\theta\implies\sin\theta=\frac W{2R}\\ H=R(1-\cos\theta)\implies\cos\theta=1-\frac HR=\frac{R-H}R $$ Using the identity $\sin^2\theta+\cos^2\theta=1$ $$\left(\frac W{2R}\right)^2+\left(\frac{R-H}R\right)^2=1$$ $$\frac{W^2}{4R^2}+\frac{R^2+H^2-2RH}{R^2}=1\\\text{ I assume you are familiar with }(a\pm b)^2=a^2+b^2\pm 2ab$$ $$\frac{W^2}4+R^2+H^2-2RH=R^2$$ $$\frac{W^2}4+H^2=2RH$$ $$R=\frac{W^2}{8H}+\frac H2$$ It is easy to see that center is at the midpoint of width. Also y-coordinate is $-R.\cos\theta$
Best Answer
Let's observe picture below. Each point $(x,y)$ of the circle that satisfy following equality can be vertice point of the rectangle :
$\left(\sqrt{(x_A-x)^2+(y_A-y)^2}\right)^2 + \left(\sqrt{(x_C-x)^2+(y_C-y)^2}\right)^2=\left(\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}\right)^2$ ,
where $A(-1,0)$ , and $C(3,-2)$ are given vertices.
For example if you choose point $B_1$ you can calculate coordinates of $D_1$ by using equalities :
$x_{D_1}=2x_O-x_{B_1}$ , and $y_{D_1}=2x_O-y_{B_1}$
Answer to sub question :
If you have length of one edge , let's say $|\bar{AB_1}|$ then you can write an extra equality :
$ |\bar{AB_1}|= \sqrt{(x_A - x)^2+(y_A-y)^2}$ , and find relation between $x$ and $y$ coordinates.