Construct the circumcircle and incircle

Try

Three noncollinear survey markers determine a circular boundary through them. A different design asks for a circle fitting inside the triangle and touching all three sides. These are different distance conditions: the circumcenter is equally distant from vertices; the incenter is equally distant from side lines. A center chosen merely because it looks central may satisfy neither.

As a worked triangle, use A=(0,0), B=(6,0), C=(0,8). Its hypotenuse has length 10. The circumcenter is the midpoint (3,4) of that hypotenuse, with radius 5. The incircle has center (2,2) and radius 2. We will justify both rather than treating those coordinates as definitions.

Predict

Predict whether a circumcenter must lie inside a triangle. Compare acute, right and obtuse sketches. Predict whether equal distances to two side lines suffice for an incircle. There are also external angle bisectors, so the word “internal” and the requirement that the center lie inside cannot be omitted.

Experiment

For any nondegenerate triangle, construct the perpendicular bisectors of AB and AC. Their unique intersection O is the circumcenter. Draw the circle centered at O through A and check B,C. Separately construct the internal bisectors at A and B; call their intersection I. Drop a perpendicular from I to AB to get radius r, then draw the circle centered at I with that radius.

Worked center	Distance to first object	Other equal distances
O=(3,4)	OA=5	OB=5 and OC=5
I=(2,2)	Distance to AB is 2	Distance to AC is 2; distance to BC is 2

For the worked triangle BC has equation 4x+3y=24. The distance from (2,2) to that line is (24−8−6)/5=2. Its contact foot is (18/5,16/5), on the actual side segment. This coordinate check corroborates the construction; it does not replace its general proof.

Understand

The perpendicular-bisector locus gives OA=OB and OA=OC. Therefore OB=OC, so O also lies on the third bisector. The first two bisectors are not parallel because the two original side lines are not parallel; hence their intersection exists uniquely. The resulting positive-radius circle passes through every vertex. Nothing here requires O to lie inside. For a right triangle it is on the hypotenuse; for an obtuse triangle it lies outside.

For an angle of measure θ strictly between 0 and 180 degrees, consider an interior point X. If ray AX splits the angle into α and β, then α+β=θ. With AX=d>0, the perpendicular distances to the full side lines are d sinα and d sinβ. This remains correct if a perpendicular foot falls behind a boundary ray: the height uses the supplementary acute angle, with the same sine. Equal distances imply sinα=sinβ, so either α=β or α+β=180 degrees. The latter is excluded by θ<180 degrees. Thus equal distances characterize the internal bisector among interior points. This branch check prevents a false assumption that every perpendicular foot lies on its forward ray.

The two internal bisectors meet inside the triangle. One way to justify this is to let the bisector from A meet the opposite side at D. Along segment AD, the signed inward distances to BA and BC vary continuously: at A the first is zero and the second positive, while at D the first is positive and the second zero. Their difference changes sign, so there is an interior point with equal distances. It lies on B’s internal bisector by the preceding locus argument; uniqueness follows from the distinct bisector lines. Equal distances to AB,AC and BC then also place I on C’s internal bisector. This uses elementary continuity of distance along a segment, not calculus.

A radius-r disk centered at I lies within each inward side half-plane, since I is distance r from each boundary line. Therefore it lies in their intersection, the whole triangle. Each perpendicular foot is a point of that disk on its side line, so it lies on the side segment, not its extension. Pythagoras shows each line has only that one point in common with the circle, proving tangency. Joining I to the vertices also gives area Δ=r(a+b+c)/2=rs, where s is semiperimeter.

Apply

For the worked triangle, area is 24 and semiperimeter is 12, so r=2 agrees with the construction. Explain why a vertex, despite zero distance to two side lines, is not an incenter: its distance to the third line is positive. In a construction report, distinguish “equal distances to vertices” from “equal perpendicular distances to lines,” and prove the contact points are on the requested segments.

Continue

Use circles next to reason about angles. Name the intercepted arc carefully; a diameter, an inscribed angle and an external pair of tangents select different geometric conditions.