Inscribed and Circumscribed Circles

Two circles for every triangle

Circle	Touches	Center	Center found from
circumscribed	all three vertices	circumcenter	perpendicular bisectors of the sides
inscribed	all three sides	incenter	bisectors of the angles

Both centers exist for every triangle, and that is a real claim — it takes three lines meeting at a single point, and three lines usually do not.

Why the perpendicular bisectors meet at one point

Points on the perpendicular bisector of $\overline{AB}$ are exactly the points equidistant from $A$ and $B$. That is the perpendicular bisector theorem and its converse.

Let $U$ be where the bisectors of $\overline{AB}$ and $\overline{BC}$ cross.

Statement	Reason
$UA = UB$	$U$ is on the perpendicular bisector of $\overline{AB}$
$UB = UC$	$U$ is on the perpendicular bisector of $\overline{BC}$
$UA = UC$	transitive property
$U$ is on the perpendicular bisector of $\overline{AC}$	converse of the perpendicular bisector theorem

So the third bisector passes through $U$ too, and $U$ is the same distance from all three vertices. A circle centered at $U$ through $A$ passes through $B$ and $C$ as well.

In the figure, $U = (5, 4)$, and each vertex is $\sqrt{20} \approx 4.47$ away.

The same argument works for the angle bisectors, with “equidistant from two sides” in place of “equidistant from two vertices”. A point on the bisector of an angle is the same distance from both of its sides, so the point where two angle bisectors cross is equidistant from all three sides.

Constructing them

Circumscribed circle.

1. Construct the perpendicular bisectors of two sides.
2. Mark where they cross. That is the circumcenter.
3. Set the compass from the circumcenter to any vertex and draw the circle.

Inscribed circle.

1. Construct the bisectors of two angles.
2. Mark where they cross. That is the incenter.
3. Construct a perpendicular from the incenter to any side. The foot of that perpendicular is where the circle touches the side.
4. Set the compass from the incenter to that foot and draw the circle.

Step 3 matters. The radius of the inscribed circle is the perpendicular distance to a side, not the distance to a vertex.

Where the circumcenter lands

Triangle	Circumcenter
acute	inside the triangle
right	at the midpoint of the hypotenuse
obtuse	outside the triangle

The right-triangle case is the angle in a semicircle read backwards: the right angle stands on a diameter, so the hypotenuse is a diameter and its midpoint is the center.

The incenter, by contrast, is always inside, since it has to sit between all three sides.

Why opposite angles of an inscribed quadrilateral are supplementary

A quadrilateral with all four vertices on a circle is inscribed in it, or cyclic.

Given: $ABCD$ is inscribed in a circle. Prove: $m\angle A + m\angle C = 180°$.

$\angle A$ is an inscribed angle standing on arc $BCD$. $\angle C$ stands on arc $BAD$. Those two arcs together make the whole circle, $360°$.

By the inscribed angle theorem, each angle is half its arc:

$$m\angle A + m\angle C = \frac{1}{2}\,m\widehat{BCD} + \frac{1}{2}\,m\widehat{BAD} = \frac{1}{2}(360°) = 180°$$

The same reasoning gives $m\angle B + m\angle D = 180°$.

Opposite angles share the circle between them, half each. A quadrilateral whose opposite angles are not supplementary cannot be inscribed in any circle — which is why a general parallelogram cannot, while a rectangle can.

Constructing a tangent from an outside point

Given a circle with center $O$ and a point $P$ outside it.

1. Draw $\overline{OP}$ and construct its midpoint $M$.
2. Draw the circle centered at $M$ through $O$ and $P$.
3. It crosses the original circle at two points, $T_1$ and $T_2$.
4. Lines $PT_1$ and $PT_2$ are the tangents.

Why it works: $\overline{OP}$ is a diameter of the new circle, and $T_1$ lies on that circle, so $\angle OT_1P$ is an angle in a semicircle — a right angle. A line through a point of the circle perpendicular to the radius there is a tangent, so $PT_1$ is tangent at $T_1$.

Worked examples

Common mistakes

Practice problems