I'm writing a program in C++ to calculate the boundary box of a circle. The user provides me with a coordinate point (10, -5) and a radius of 23.
I'm not very good at geometry and I need help in calculating the bounding box of this circle. How would I calculate this?
[Math] How to calculate a bounding box of a circle
circlesgeometry
Related Solutions
By scaling one can assume that $R_a=1$, and we can put $R_b=r$. This answer only looks at the case where the second circle of radius $r$ is positioned so that its point of tangency with the unit circle (center at $A=(0,0)$ lies in the first quadrant. We also assume that the $r$ circle protudes enough in the $x$ direction so that the bbox needs to extend to the right of $x=1$, and that the $r$ circle is far enough down from the top that the entire $r$ circle lies below the line $y=1$.
The center of the $r$ circle lies at $((1+r) \cos \theta, (1+r) \sin \theta)$ where $\theta$ is the angle formed by the center of the $r$ circle, the origin, and the positive $x$ axis. Then the condition that the $r$ circle lies below $y=1$ is $$[1]\ \ \ r+(1+r)\sin \theta \le 1.$$ And the condition that the $r$ circle protrudes beyond the line $x=1$ is $$[2]\ \ \ r+(1+r)\cos \theta \ge 1.$$ The ratio $k$ under these assumptions comes from the fact that, in this situation, the height of the bbox is $2$ while its width, extending from $x=-1$ to the rightmost point of the $r$ circle, is $1+(1+r)\cos \theta +r$. This gives the ratio as $$k = \frac{2}{(1+r)(1+\cos \theta)},$$ which solves for $\cos \theta$ as $$\cos \theta=\frac{2}{k(1+r)}-1.$$ One can then find $\theta$ via arccos. Even in this simple case, it is necessary to know that [1] and [2] are satisfied, in order for the expression for $\cos \theta$ to achieve $k$ be valid. I've attempted to eliminate $\cos \theta$ and $\sin \theta$ from conditions [1] and [2], but it gets involved.
Note that here the value of $k$ must lie in the interval $[1/(1+r),1]$ in order for there to be a possible $r$ circle, where the smallest $k$ corresponds to the case where the $r$ circle is tangent to the unit circle at $(1,0)$. One naturally also needs $r<1$ or else the $r$ circle will extend beyond $y=1$ and change the formula for $k$.
It remains to look at other cases, for example the case wherein the $r$ circle extends both to the right of $x=1$ and above $y=1$. And the conditions [1] and [2] need to be restated without $\theta$, for a complete answer. There may be another way to look at this question, which would not involve so many cases and conditions.
So you know the distance $d=CD$ between center and boundary. Then you can write
$$\cos\angle ECD = \tfrac dr \qquad \angle ECF = 2\angle ECD = 2\arccos\tfrac dr$$
Now the length of an arc is $r$ times its angle, so the outside arc is $r\angle ECF$ and the inside arc is
$$s = r(2\pi-\angle ECF)=2r(\pi-\arccos\tfrac dr)$$
You want that number to be equal to some given value, so you want to solve the above equation for $r$. Unfortunately, that equation is transcendental, so you can't expect a closed form solution to your problem. Your best bet is some form of iterative numeric approximation.
As you can see from that plot, you can expect that for many possible ratios of $\frac sd$, you get two distinct solutions for $r$.
The apex (with the vertical tangent) appears to be at
$$ s/d\approx 5.94338774142760424162091392488776998544210982523814509283191138267355981 \\ r/d\approx 1.06193134974748196175464922830803488867448733227482933642882008697882597 $$
Best Answer
By bounding box I'm assuming you mean the box (square) in which the circle is inscribed, like in this picture:
Notice that the radius of the circle is exactly half the length of a side of the square.
So if the center of the circle is $(10,-5)$ and the radius of the circle is $23$, and if we're assuming a standard coordinate system ($y$-values increase in the up direction and $x$-values increase in the right direction), then the corners of the box are located at the following points:
\begin{align*} \text{upper left corner} &= (10-23, -5+23)\\ &= (-13, 18)\\[0.3cm] \text{upper right corner} &= (10+23, -5+23)\\ &= (33, 18)\\[0.3cm] \text{lower right corner} &= (10+23, -5-23)\\ &= (33, -28)\\[0.3cm] \text{lower left corner} &= (10-23, -5-23)\\ &= (-13, -28) \end{align*}