Circles show up on every competition. Know the basic formulas, the inscribed angle theorem,
and the tangent-radius relationship, and you can solve most circle problems. The key idea:
draw radii — they create isosceles triangles and right angles with tangents.
📚 Key Concepts
Circle Terminology
Radius (r): distance from center to any point on the circle
Diameter (d): $ d = 2r $ — longest chord, passes through center
Chord: line segment with both endpoints on the circle
Arc: portion of the circumference between two points
Central angle: angle with vertex at the center, sides are radii
Inscribed angle: angle with vertex on the circle, sides are chords
Tangent: line that touches the circle at exactly one point
Sector: "pizza slice" — region bounded by two radii and an arc
Segment: region bounded by a chord and an arc
Circumference & Area
Circumference (perimeter of a circle):
\[
C = 2\pi r = \pi d
\]
Area:
\[
A = \pi r^2
\]
Arc Length & Sector Area
For a sector with central angle $\theta$ (in degrees):
Arc length: fraction of the full circumference
\[
\text{Arc length} = \frac{\theta}{360°} \times 2\pi r
\]
Corollary: inscribed angles that intercept the same arc are equal.
Thales' Theorem: An angle inscribed in a semicircle is a right angle (90°).
If you see a triangle inscribed with the diameter as one side, it's a right triangle!
Chord Properties
The perpendicular from the center to a chord bisects the chord (and its arc)
Converse: the line from center to midpoint of a chord is perpendicular to the chord
Equal chords are equidistant from the center
Chord length formula: $ L = 2\sqrt{r^2 - d^2} $ where $d$ = distance from center to chord
This is a classic Pythagorean theorem setup: radius, half-chord, and distance form a right triangle.
Tangent Properties
Radius ⊥ tangent: A radius is perpendicular to the tangent at the point of contact
Two tangents from a point: Two tangent segments from the same external point are equal in length
The line from the external point to the center bisects the angle between the two tangents
Always draw the radius to the tangent point — it creates a right angle, and right triangles mean Pythagoras!
Tangent-Chord Angle Theorem
The angle between a tangent and a chord equals half the measure of the intercepted arc (the arc "inside" the angle).
Also called the "alternate segment theorem" in some curricula.
⚠️ Common Mistake: Don't confuse central angles and inscribed angles. Central angles equal the arc;
inscribed angles are half the arc. Also: arc length uses $2\pi r$ (circumference), sector area uses $\pi r^2$ (area).
Don't mix up the formulas — both have the $\frac{\theta}{360°}$ fraction, but the base formula is different.
💡 Key Insight: In circle problems, your first move should almost always be: draw the radii.
Radii create isosceles triangles (two sides equal) and right angles with tangents. If you see a tangent, draw the
radius to the point of tangency — you get a right triangle and can use Pythagoras. If you see a chord and the center,
drop a perpendicular — another right triangle. Radii unlock everything.
✏️ Example Problems
📝 Example 1 (Arc length & sector area)
A circle of radius 6 has a sector with central angle 60°.
What is the area of the sector? (Leave $\pi$ out of your answer — just give the coefficient.)
Step 1: Full area of the circle.
$ A_{full} = \pi r^2 = \pi (6)^2 = 36\pi $
Step 2: Find what fraction of the circle the sector is.
Triangle ABC is inscribed in a circle with AC as the diameter.
If angle BAC = 35°, what is the measure of angle BCA (in degrees)?
Step 1: Apply Thales' theorem.
Since AC is the diameter, angle ABC (the angle subtended by the diameter) = 90°.
So triangle ABC is a right triangle with right angle at B.
Step 2: Angles in a triangle sum to 180°.
$ \angle BAC + \angle ABC + \angle BCA = 180° $
$ 35° + 90° + \angle BCA = 180° $
$ 125° + \angle BCA = 180° $
$ \angle BCA = 55° $
Answer: 55
📝 Example 3 (Tangent + Pythagoras)
From a point P outside a circle of radius 5, a tangent is drawn touching the circle at T.
If the distance from P to the center O is 13, what is the length of tangent PT?
Step 1: Draw radius OT to the point of tangency.
OT = radius = 5
Step 2: Radius is perpendicular to tangent.
$ OT \perp PT $, so triangle OTP is a right triangle (right angle at T).
Step 3: Apply Pythagorean theorem.
Hypotenuse OP = 13, one leg OT = 5, find other leg PT.