Circumcenter
Three points not on one line. Is there always a circle through all three, and where is its center?
Where you are 4 in Triangle centers
Why this exists
Pick three points that are not on one line. Is there a circle that passes through all three? Someone holding three pieces of a broken plate wants to know where the plate’s center was. Someone choosing a site for a radio mast that must be the same distance from three towns is asking the same thing. Both want one point that is equally far from three given points.
For two points the answer is already known. By the perpendicular bisector theorem, the points equally far from A and B are exactly the points of the perpendicular bisector of \overline{AB}, a locus that is a straight line. Three points give three such lines, one for each pair. Two lines that are not parallel always meet, so two of the three conditions can always be met at once. The real question is whether the third line passes through that same point.
It always does, as the perpendicular bisector lesson proves. That point is the circumcenter, and it settles the plate, the mast and every other “equally far from three points” question in one move. This lesson is about what that point does: the circle it carries, where it sits for each shape of triangle, and how to find it.
Definition
Three or more lines are concurrent when they all pass through one point. That point is their point of concurrency. Two lines that are not parallel always meet, so the word only has bite for three or more lines: concurrency says that a third line goes through the point the first two already share.
The perpendicular bisector of a side of a triangle is the line through the midpoint of that side and perpendicular to it (Perpendicular bisector of a segment). Every triangle has three, one for each side.
The circumcenter of a triangle is the point where the three perpendicular bisectors of its sides meet. It is usually called O. (British texts often spell it circumcenter.)
The circumscribed circle, or circumcircle, of a triangle is the circle that passes through all three of its vertices. Its center is the circumcenter, and its radius, written R, is the circumradius. The triangle is then said to be inscribed in the circle.
That the three bisectors really do meet in one point is a theorem, not part of the definition. It is the first result below.
Intuition
Think of each perpendicular bisector as a line of points that are fair to two vertices: every point of \ell is as far from A as from B, every point of m as far from A as from C, and every point of n as far from B as from C. The point O where all three cross is fair to every pair at once.
Once O is the same distance R from all three vertices, the circle with center O and radius R runs through A, B and C. A set of points all the same distance from one center is what a circle is.
One feature of the picture surprises people: nothing forces O to be inside the triangle. Flatten the triangle until one angle opens past 90^\circ and the bisectors still meet, but out beyond the longest side.
Results
Circumcenter theorem. The perpendicular bisectors of the three sides of any triangle are concurrent. Their common point O is the same distance from all three vertices, and it is the only point that is:
The circumscribed circle. Through any three points that are not on one line there is exactly one circle. Its center is the circumcenter of the triangle the points form, and its radius is R.
Where the circumcenter sits. In an acute triangle the circumcenter is inside the triangle. In a right triangle it is the midpoint of the hypotenuse. In an obtuse triangle it is outside, across the side opposite the obtuse angle.
Right triangles. The hypotenuse of a right triangle is a diameter of its circumcircle, so
where c is the length of the hypotenuse.
Angle at the center. When O is inside \triangle ABC,
and the same holds for the other two sides. This is a first case of the inscribed angle theorem.
Finding O in coordinates. Write two of the perpendicular bisectors as equations and solve them together; the third is then automatic and makes a free check. An equivalent route sets squared distances equal, PA^2 = PB^2 and PA^2 = PC^2. The squares of x and y cancel, leaving two linear equations, and each of them is a perpendicular bisector written without slopes.
Why it is true
The bisectors are concurrent. This is proved in full in the perpendicular bisector lesson: two of the bisectors cannot be parallel, so they cross at a point O; that point is equally far from all three vertices, so the converse of the theorem puts it on the third bisector too; and any point equally far from all three lies on two crossing lines, so O is the only one.
There is exactly one circle. The circle with center O and radius OA passes through all three vertices, because OB and OC both equal OA. Any circle through A, B and C has a center equally far from all three, and O is the only such point, so its center is O and its radius is OA.
Where it sits. Join O to the three vertices. Because OA = OB = OC, this makes three isosceles triangles, and the base angles of an isosceles triangle are equal (Isosceles and equilateral triangles). Call the two base angles of \triangle OBC both p, those of \triangle OCA both q, and those of \triangle OAB both r.
When O is inside. Each angle of the triangle is split, by the segment to O, into two base angles:
The angles of \triangle ABC add to 180^\circ (Triangle angle sum), so 2(p + q + r) = 180^\circ, that is, p + q + r = 90^\circ. Then m\angle C = 90^\circ - r, which is less than 90^\circ, and in the same way \angle A and \angle B are less than 90^\circ. Every angle is acute. The angle sum of \triangle OAB also gives m\angle AOB = 180^\circ - 2r = 2(90^\circ - r) = 2\,m\angle ACB, which is the angle-at-the-center result.
When O is on a side. O cannot be a vertex, because OA = 0 would force OB = OC = 0 as well. O cannot lie on a side’s line outside the side either: on line AB, the only point with OA = OB is the midpoint. So say O lies on \overline{AB}; since OA = OB, it is the midpoint. Now \triangle OAB has collapsed onto a segment, and only \triangle OBC and \triangle OCA remain, so m\angle A = q, m\angle B = p and m\angle C = p + q. The angle sum gives 2(p + q) = 180^\circ, so m\angle C = 90^\circ.
When O is outside. Then O is on the far side of some side’s line from the opposite vertex (it is on none of the three lines, as just shown). Name the vertices so that O and C are on opposite sides of line AB. Then \overline{CO} crosses line AB at a point X, with X strictly between C and O.
The first job is to show that X lies strictly between A and B. Let M be the midpoint of \overline{AB}. O is on the perpendicular bisector of \overline{AB}, so \overline{OM} is perpendicular to line AB, and for any point P of line AB the Pythagorean theorem in the right triangle OMP gives OP^2 = OM^2 + MP^2 (true also when P = M). Compare X with A:
- X lies on \overline{OC} and is not C, so OX < OC = R = OA;
- so OM^2 + MX^2 < OM^2 + MA^2, which gives MX < MA.
M is the midpoint and MA = MB, so the points of line AB closer to M than MA are exactly the points strictly between A and B. X is one of them.
That has two consequences. Ray CO meets \overline{AB} at an inner point, so it runs inside \angle ACB and m\angle C = p + q as before. And ray AB passes through X, an inner point of \overline{CO}, so it runs inside \angle CAO: the angle of \triangle OAB at A is now taken away instead of added, m\angle A = q - r, and in the same way m\angle B = p - r. The angle sum gives 2(p + q) - 2r = 180^\circ, so
which is more than 90^\circ. The obtuse angle is \angle C, opposite the side \overline{AB} that O lies across.
Reading it backwards. A triangle is exactly one of acute, right and obtuse, and O is exactly one of inside, on a side and outside. The three cases send the three positions to three different kinds of triangle, so each kind can only come from its own position. An acute triangle cannot have O on a side (that would make it right) or outside (that would make it obtuse), so O is inside; the other two kinds go the same way. In a right triangle, O is the midpoint of the side opposite the right angle, which is the hypotenuse, so R = OA = \tfrac{1}{2}c.
Worked examples
Testing three lines for concurrency
Are the lines y = x + 1, y = -x + 5 and y = 3x - 3 concurrent?
SolutionLines with different slopes always meet, so find where the first two meet and then ask whether the third passes through that point. A common point has one y-value, so set the first two right-hand sides equal:
and then y = 2 + 1 = 3. The first two lines meet at (2, 3). Substituting x = 2 into the third line gives y = 3(2) - 3 = 3, so (2, 3) is on it as well. The three lines are concurrent, and their point of concurrency is (2, 3).
Had the third line been y = 3x - 2, it would give y = 4 at x = 2 and miss the point. The three lines would then not be concurrent; they would enclose a small triangle instead.
The circumradius of an isosceles triangle
In \triangle ABC, AB = AC = 10 and BC = 12. Find the circumradius.
SolutionLet M be the midpoint of \overline{BC}. Since AB = AC, the point A is on the perpendicular bisector of \overline{BC}, and so is M; that bisector is the line AM. The circumcenter O lies on it, so the whole problem lives on one line.
BM = 6 and \angle AMB is a right angle, so \displaystyle AM = \sqrt{10^2 - 6^2} = 8. Put M at the origin with A(0, 8), B(-6, 0), C(6, 0), and write O(0, y). Setting OB^2 = OA^2:
So O is \tfrac{7}{4} above M, between M and A, and
Check with the other vertex: OB^2 = 36 + \tfrac{49}{16} = \tfrac{625}{16}, and \left(\tfrac{25}{4}\right)^2 = \tfrac{625}{16}. OC = OB comes free by symmetry. Since y > 0, the center is inside the triangle, so by the Results this triangle is acute.
A right triangle on the axes
Find the circumcenter and circumradius of the triangle with vertices A(0, 0), B(8, 0) and C(0, 6).
SolutionSide \overline{AB} lies along the x-axis with midpoint (4, 0), so its perpendicular bisector is the vertical line x = 4. Side \overline{AC} lies along the y-axis with midpoint (0, 3), so its perpendicular bisector is the horizontal line y = 3. They meet at O(4, 3).
The angle at A, between the two axes, is a right angle, and just as the Results predict, O is the midpoint of the hypotenuse \overline{BC}: \left(\tfrac{8 + 0}{2}, \tfrac{0 + 6}{2}\right) = (4, 3). With the The distance formula,
The circumradius is R = 5, half of the hypotenuse \displaystyle BC = \sqrt{8^2 + 6^2} = 10, as the right-triangle result says.
An acute triangle with a fractional center
Find the circumcenter and circumradius of the triangle with vertices A(0, 0), B(10, 0) and C(2, 6).
SolutionStart with a horizontal side if there is one, because its bisector is the simplest possible line. Side \overline{AB} has midpoint (5, 0), so its perpendicular bisector is x = 5.
Side \overline{AC} has midpoint (1, 3) and slope \tfrac{6 - 0}{2 - 0} = 3. A line perpendicular to it has slope -\tfrac{1}{3}, the negative reciprocal (Slopes of perpendicular lines), so its perpendicular bisector is
O is on both bisectors, so substitute x = 5: y = 3 - \tfrac{4}{3} = \tfrac{5}{3}. The circumcenter is O\left(5, \tfrac{5}{3}\right).
Check all three distances, squared so that no roots appear until the end:
So \displaystyle R = \sqrt{\tfrac{250}{9}} = \tfrac{5\sqrt{10}}{3} \approx 5.27.
O is inside the triangle: it is above \overline{AB}, below line AC (which is y = 3x, at height 15 when x = 5) and below line BC (which is y = \tfrac{3}{4}(10 - x), at height 3.75 when x = 5). By the Results, every angle of this triangle is therefore acute.
An obtuse triangle: the center outside
Find the circumcenter of the triangle with vertices A(0, 0), B(8, 0) and C(10, 4), and say where it lies.
SolutionSide \overline{AB} is horizontal with midpoint (4, 0), so its perpendicular bisector is x = 4. Side \overline{BC} has midpoint (9, 2) and slope \tfrac{4 - 0}{10 - 8} = 2, so its perpendicular bisector has slope -\tfrac{1}{2}:
At x = 4 this gives y = 2 + \tfrac{5}{2} = \tfrac{9}{2}, so the circumcenter is O\left(4, \tfrac{9}{2}\right). Checking the squared distances:
so \displaystyle R = \tfrac{\sqrt{145}}{2} \approx 6.02.
Line AC is y = \tfrac{2}{5}x, at height \tfrac{8}{5} when x = 4. O is at height \tfrac{9}{2}, above the line, while B is below it. So O and B are on opposite sides of line AC: the circumcenter is outside the triangle, across \overline{AC}. By the Results the angle opposite \overline{AC}, which is \angle B, must be obtuse. The picture agrees: C is further right than B, so ray BC leans right while ray BA points straight left, and the angle between them is more than a right angle.
A mast equally far from three towns
Three towns sit at A(4, 5), B(-4, 1) and C(5, -2) on a map marked in kilometers. Where should a radio mast go so that it is the same distance from all three, and how far is that?
SolutionThe mast must stand at the circumcenter P(x, y). No side here is horizontal or vertical, so instead of working with slopes, write the condition PA = PB directly, squared so that there are no roots:
Expanding, x^2 and y^2 appear on both sides and cancel:
which simplifies to -16x - 8y + 24 = 0, that is, 2x + y = 3. This linear equation is the perpendicular bisector of \overline{AB}. Do the same with PA = PC:
so 2x - 14y + 12 = 0, that is, x - 7y = -6.
Now solve the two together. The first gives y = 3 - 2x, and substituting into the second, x - 7(3 - 2x) = -6, so 15x = 15 and x = 1, then y = 1. The mast goes at (1, 1).
Every distance is 5: \displaystyle PA = \sqrt{3^2 + 4^2} = 5, \displaystyle PB = \sqrt{5^2 + 0^2} = 5 and \displaystyle PC = \sqrt{4^2 + 3^2} = 5. The mast is 5 km from each town. The condition PB = PC was never used, and it holds anyway, which is the circumcenter theorem at work.
Angles from the center
Point O is the circumcenter of acute \triangle ABC, with m\angle OAB = 20^\circ and m\angle OBC = 40^\circ. Find the three angles of the triangle and m\angle AOB.
SolutionThe triangle is acute, so O is inside it and the bookkeeping from Why it is true applies. OA = OB, so \triangle OAB is isosceles and its base angles are equal: m\angle OBA = m\angle OAB = 20^\circ. This is r. In the same way OB = OC gives m\angle OCB = m\angle OBC = 40^\circ, which is p. Since p + q + r = 90^\circ, the third pair is q = 30^\circ, so m\angle OCA = m\angle OAC = 30^\circ. Adding the two pieces at each vertex:
They add to 180^\circ, as they must. Finally, the angle sum of \triangle OAB gives m\angle AOB = 180^\circ - 20^\circ - 20^\circ = 140^\circ, which is twice m\angle ACB = 70^\circ, as the angle-at-the-center result says.
Boundaries
Three points on one line have no circumcenter. The perpendicular bisectors of \overline{AB} and \overline{BC} are then both perpendicular to the same line, so they are parallel, and they cross it at different midpoints, so they never meet. No point is equally far from all three, and no circle passes through them. Before it breaks, it stretches: for A(-1, 0), B(1, 0) and C(0, h) the circumcenter is \left(0, \tfrac{h^2 - 1}{2h}\right), and as h shrinks toward 0 this point runs off without bound and the circle grows without bound.
A fourth point usually misses. Any three points not on a line have their circle, but a fourth point is on it only by luck or by design. Quadrilaterals whose four vertices do share a circle are special, and cyclic quadrilaterals are about exactly which ones.
“Center” means the center of the circle, not the middle of the triangle. For an obtuse triangle the circumcenter is outside the triangle altogether. Being equally far from the vertices also says nothing about distances to the sides, which is what the Incenter measures, and nothing about where the triangle balances, which is the Centroid.
The plane matters. In space, the points equally far from two points form a plane rather than a line, and the points equally far from three points form a whole line, through O and perpendicular to the plane of the triangle. A single center is a fact about the flat plane.
Where this goes next
The circumcenter is the first of the four classic triangle centers, and the proof pattern, two lines meet and the third is forced through the same point, comes back for the Incenter, where angle bisectors and distances to sides replace perpendicular bisectors and distances to vertices, and for the Centroid and Orthocenter and the Euler line. The orthocenter lesson also shows the circumcenter, centroid and orthocenter lying on one line.
With compass and straightedge, the same two bisectors become a construction: Constructions with circles draws the circle through three given points. In coordinates, the center and radius found here are exactly what Equation of a circle needs to write the circle’s equation.
The angle-at-the-center result, m\angle AOB = 2\,m\angle ACB, is the seed of the inscribed angle theorem, and the right-triangle case, a hypotenuse that is a diameter, is the special case where the inscribed angle is 90^\circ. Asking when a fourth point lands on the circle leads to cyclic quadrilaterals.
Exercises
Concurrent lines
- Are the lines y = 2x - 1, y = -x + 5 and y = x + 1 concurrent? If so, name the point of concurrency.
- Are the lines y = x + 2, y = -2x + 8 and y = 3x - 1 concurrent?
- Find the value of k that makes the lines y = 2x + 1, y = -x + 7 and y = kx - 4 concurrent.
Distances from the circumcenter
- O is the circumcenter of \triangle ABC, with OA = 3x + 4 and OC = 7x - 12. Find x and OB.
- O is the circumcenter of \triangle DEF, with OD = x^2 - 5 and OE = 4x. Find x and the circumradius.
- O is the circumcenter of \triangle PQR and M is the midpoint of \overline{QR}. If QR = 24 and OM = 5, find OQ.
- A triangle has circumradius 10 and one side of length 12. How far is the circumcenter from the line containing that side?
Right triangles and where the center sits
- A right triangle has legs 5 and 12. Find its circumradius.
- A triangle has sides 8, 15 and 17. Show that it is a right triangle, say where its circumcenter is, and find the circumradius.
- Say whether the circumcenter lies inside, on a side of, or outside a triangle with angles (a) 35^\circ, 70^\circ, 75^\circ; (b) 25^\circ, 65^\circ, 90^\circ; (c) 20^\circ, 40^\circ, 120^\circ.
Coordinates
- Find the circumcenter and circumradius of the triangle with vertices A(0, 0), B(6, 0) and C(2, 4).
- Find the circumcenter of the triangle with vertices A(-3, 2), B(5, 2) and C(1, -2). What does its position tell you about \angle C?
- Find the circumcenter of the triangle with vertices A(0, 0), B(4, 0) and C(6, 4), and say whether it is inside or outside the triangle.
- Find the circumcenter and circumradius of the triangle with vertices A(5, 7), B(-2, 6) and C(2, -2).
- Three wells sit at (1, 7), (7, -1) and (-5, -1) on a map marked in kilometers. A pump is to be placed the same distance from all three. Where does it go, and how far is it from each well?
- (a) Is there a circle through all four of the points (0, 0), (6, 0), (0, 8) and (6, 8)? (b) Through all four of (0, 0), (6, 0), (0, 8) and (7, 6)?
Angles at the circumcenter
- O is the circumcenter of acute \triangle ABC, with m\angle OAB = 30^\circ and m\angle OBC = 25^\circ. Find the three angles of the triangle.
- O is the circumcenter of acute \triangle ABC and m\angle ACB = 50^\circ. Find m\angle OAB and m\angle AOB.
- O is the circumcenter of acute \triangle ABC and m\angle AOB = 130^\circ. Find m\angle ACB.
Reasoning
- \overline{AB} is a diameter of a circle, and C is any other point on the circle. Use the results of this lesson to explain why \angle ACB is a right angle.
- Can the circumcenter of a triangle be one of its vertices? Explain.
- In \triangle ABC, M is the midpoint of \overline{BC}, and the circumcenter lies on line AM. Show that AB = AC or \angle A is a right angle, and give a triangle where the second happens without the first.
- A triangle has circumradius 5. Can one of its sides be 12 long? Show that no side of a triangle is longer than 2R, and say when a side equals 2R.
- The circumcenter of a triangle lies outside it. Explain why the side that the circumcenter lies across is the longest side of the triangle.
Answers
- Concurrent at (2, 3). The first two meet where 2x - 1 = -x + 5, so x = 2 and y = 3. The third line at x = 2 gives y = 2 + 1 = 3, the same point.
- Not concurrent. The first two meet where x + 2 = -2x + 8, so x = 2 and y = 4. The third line at x = 2 gives y = 5 \ne 4.
- k = \tfrac{9}{2}. The first two meet where 2x + 1 = -x + 7, so x = 2 and y = 5. The third must pass through (2, 5): 5 = 2k - 4, so 2k = 9.
- x = 4 and OB = 16. OA = OC gives 3x + 4 = 7x - 12, so 16 = 4x. Then OA = 3(4) + 4 = 16, and OB = OA.
- x = 5 and radius 20. OD = OE gives x^2 - 5 = 4x, so x^2 - 4x - 5 = 0 and (x - 5)(x + 1) = 0. The root x = -1 makes OE = -4, impossible for a distance, so x = 5, OE = 20 and OD = 25 - 5 = 20.
- OQ = 13. O is on the perpendicular bisector of \overline{QR}, which meets it at M at a right angle, so \triangle OMQ has a right angle at M with QM = 12: \displaystyle OQ = \sqrt{5^2 + 12^2} = 13.
- 8. As in exercise 6, the perpendicular from O meets the side at its midpoint M, making a right triangle with hypotenuse R = 10 and leg 6: \displaystyle OM = \sqrt{10^2 - 6^2} = 8.
- R = \tfrac{13}{2}. The hypotenuse is \displaystyle \sqrt{5^2 + 12^2} = 13, and the circumradius of a right triangle is half the hypotenuse.
- 8^2 + 15^2 = 64 + 225 = 289 = 17^2, so by the converse of the Pythagorean theorem the angle opposite the side of length 17 is a right angle. The circumcenter is the midpoint of that side, and R = \tfrac{17}{2}.
- (a) Inside, since all three angles are acute. (b) On a side: the midpoint of the side opposite the 90^\circ angle. (c) Outside, across the side opposite the 120^\circ angle.
- O(3, 1) and \displaystyle R = \sqrt{10}. The bisector of \overline{AB} is x = 3. \overline{AC} has midpoint (1, 2) and slope 2, so its bisector is y - 2 = -\tfrac{1}{2}(x - 1), and at x = 3, y = 1. Check: OA^2 = 9 + 1 = 10, OB^2 = 9 + 1 = 10, OC^2 = 1 + 9 = 10.
- O(1, 2), the midpoint of \overline{AB}, so \angle C is a right angle. \overline{AB} is horizontal with midpoint (1, 2), so its bisector is x = 1. On that line OA^2 = 16 + (y - 2)^2 and OC^2 = (y + 2)^2; setting them equal and canceling y^2 gives 16 - 4y + 4 = 4y + 4, so y = 2. Here R = 4.
- O\left(2, \tfrac{7}{2}\right), outside the triangle. The bisector of \overline{AB} is x = 2. \overline{BC} has midpoint (5, 2) and slope 2, so its bisector is y - 2 = -\tfrac{1}{2}(x - 5), and at x = 2, y = \tfrac{7}{2}. Line AC is y = \tfrac{2}{3}x, at height \tfrac{4}{3} when x = 2, so O is above it and B is below: O lies across \overline{AC}, and \angle B is obtuse. The circumradius is \displaystyle \tfrac{\sqrt{65}}{2}.
- O(2, 3) and R = 5. PA^2 = PB^2 expands to -10x - 14y + 74 = 4x - 12y + 40, that is, 7x + y = 17. PA^2 = PC^2 expands to -10x - 14y + 74 = -4x + 4y + 8, that is, x + 3y = 11. Substituting y = 17 - 7x gives x + 51 - 21x = 11, so x = 2 and y = 3. Check: \displaystyle OA = \sqrt{3^2 + 4^2} = 5.
- At \left(1, \tfrac{3}{4}\right), which is \tfrac{25}{4} = 6.25 km from each well. The wells at (7, -1) and (-5, -1) give the bisector x = 1. On it, the distance to (1, 7) is 7 - y and the distance to (7, -1) is \displaystyle \sqrt{36 + (y + 1)^2}. Squaring and setting them equal, 49 - 14y = 37 + 2y, so y = \tfrac{3}{4}, and the distance is 7 - \tfrac{3}{4}.
- (a) Yes. (b) No. The first three points make a right angle at (0, 0), so their circle has its center at the midpoint of the hypotenuse, (3, 4), and radius 5. In (a), (6, 8) is \displaystyle \sqrt{3^2 + 4^2} = 5 from the center, so it is on the circle. In (b), (7, 6) is \displaystyle \sqrt{4^2 + 2^2} = \sqrt{20} from it, not 5. The circle through the first three points is the only one, so no circle passes through all four.
- m\angle A = 65^\circ, m\angle B = 55^\circ, m\angle C = 60^\circ. Here r = 30^\circ and p = 25^\circ, so q = 90^\circ - 30^\circ - 25^\circ = 35^\circ. Then m\angle A = q + r, m\angle B = r + p and m\angle C = p + q.
- m\angle OAB = 40^\circ and m\angle AOB = 100^\circ. Since m\angle C = 90^\circ - r, r = 90^\circ - 50^\circ = 40^\circ, and the angle sum of \triangle OAB gives m\angle AOB = 180^\circ - 2(40^\circ).
- 65^\circ. \triangle OAB is isosceles, so each base angle is r = \tfrac{1}{2}(180^\circ - 130^\circ) = 25^\circ, and m\angle ACB = 90^\circ - r. It is half of m\angle AOB, as the angle-at-the-center result says.
- C is not on line AB, since a line meets a circle in at most two points and those are A and B, so A, B, C form a triangle. There is only one circle through three such points, so the given circle is the circumcircle of \triangle ABC and its center O is the circumcenter. O is the midpoint of the diameter \overline{AB}, so the circumcenter lies on a side, and the on-a-side case of Why it is true gives m\angle ACB = 90^\circ.
- No. If O = A then OA = 0, and OB = OC = OA = 0 would force B and C to be the point A too. A triangle has three different vertices.
- Call the circumcenter O. If O = M, the circumcenter is on side \overline{BC}, so \angle A is a right angle. Otherwise O and M are two different points of the perpendicular bisector of \overline{BC} (O by definition, M because it is the midpoint), so that bisector is line OM, which is line AM. Then A is on the perpendicular bisector of \overline{BC}, and the perpendicular bisector theorem gives AB = AC. For the second without the first, take A(0, 0), B(4, 0), C(0, 2): the right angle is at A, O = M = (2, 1) is on line AM, and AB = 4 \ne 2 = AC.
- No: every side is at most 2R = 10. Take a side \overline{PQ} with midpoint M. If O = M, then PQ = 2\,OP = 2R. Otherwise \triangle OMP has a right angle at M and hypotenuse OP = R, and a leg is shorter than the hypotenuse (MP^2 = R^2 - OM^2 < R^2), so PQ = 2\,MP < 2R. A side equals 2R exactly when the circumcenter is its midpoint, that is, when it is the hypotenuse of a right triangle.
- Name the vertices so that O lies across \overline{AB} from C. The bookkeeping in Why it is true gives m\angle C = 90^\circ + r > 90^\circ, so \angle C is obtuse. The other two angles then add to less than 90^\circ, so \angle C is the largest angle, and the longest side is the one opposite the largest angle (Sides and angles in order). That side is \overline{AB}.