Let me begin by apologizing for the recent infrequency of site updates and additions, in particular on the subject of the minimum enclosing circle of finitely many points in the plane. This post is long overdue.

I first want to comment some more on computing the radius of the circumscribed circle of a triangle , where , , and are the vertices. It turns out that the diameter of the circumscribed triangle is length of any side divided by the sine of the angle opposite that side. Note that by the law of sines, we only need to prove

,

where we abuse notation and label the angle (in radians) formed by sides as .

We break down the proof into three cases based on whether the triangle is acute, right, or obtuse. The proof for a right triangle is fairly straightforward, and I leave it to the reader.

Suppose that is acute. Denote the circumcenter by . Consider the isosceles triangle . If we let denote the midpoint of side , then we have . I claim that the common value of these angles is . Indeed, define real numbers by

Since the triangles and are also isosceles, we have

Recalling that the sum of the angles of a triangle is , we see that , and therefore

The length of the sides is , therefore

Now suppose that is obtuse. Relabeling if necessary, we may assume that angle is obtuse. I claim that angle . Consider the isosceles triangles and . Observe that and

and ,

Since , we conclude that . Thus,

I now want to turn to the question of when the circumscribed circle coincides with the smallest enclosing circle of . Not surprisingly, the answer hinges on whether is acute, right, or obtuse.

Theorem.If is acute or right, then the circumscribed circle is the smallest enclosing circle. If is obtuse, then the circumscribed circle is strictly larger than the smallest enclosing circle. Moreover, the smallest enclosing circle is centered at the midpoint of the longest side with radius half the length of this side.

*Proof. *In all cases, let and denote the center and radius, respectively, of the circumscribed circle. A week or so ago, we proved the following classification:

- is inside the triangle if and only if is acute;
- is the midpoint of the hypotenuse if and only if is right;
- is outside the triangle if and only if is obtuse.

Also recall that the minimum enclosing disk centered at with radius must contain at least two points of on its boundary. Without loss of generality, assume that lie on the boundary. Since is equidistant from and , lies on the perpendicular bisector of side .

Suppose is acute. If , then since , we have that . Looking at the orthogonal projection of onto the perpendicular bisector of , we see that . But then the point is not contained in the minimum enclosing disk, which is a contradiction.

Suppose is right. If , then it is evident from the Pythagorean theorem that or .

Now suppose that is obtuse. Since lies outside the triangle, it is clear that and . Without loss of generality, suppose that is the obtuse angle, so that is the longest side. If does not lie on the perpendicular bisector of , then either or . If lies on the perpendicular bisector to but , then

by the Pythagorean theorem.