Circles
Chords, tangents, inscribed angles, power of a point, arc length, sectors, inscribed/circumscribed circles
What you need to know
- Inscribed Angle Theorem. An inscribed angle (corner on the circle) is half the arc it cuts off. So an inscribed angle whose sides pass through the two ends of a diameter is a right angle. Example: an inscribed angle on a arc is .
- Tangents. A tangent line is perpendicular to the radius at the point where it touches. Two Tangent Theorem: the two tangent segments from a point outside the circle are equal. Example: from a point away from the center of a circle with radius , both tangents are long.
- Power of a Point. If chords and cross at , then . From a point outside the circle, with a tangent and a line (a secant) meeting the circle at and then : . Example: chords with , , give . A secant with , gives .
- Arcs and sectors. A central angle of degrees cuts off of the circle: the arc length is , and the sector (pizza slice) has area . Example: with and , the arc is long and the sector's area is .
How AMC 10 tests it
- Shaded regions (problems 3–12): circles in squares, or semicircles on the sides of a triangle. Subtract sectors and triangles.
- Tangent lines from a point, or one line tangent to two circles. Radii to the points of tangency give a right triangle or a trapezoid with two right angles.
- Problems 17–23: Power of a Point for a length, or a rope wrapping around a polygon, giving a sum of sectors.
Standard approaches
- Draw the radius to every point of tangency and every chord end. Look for right angles, and for isosceles triangles (radii are equal).
- Turn inscribed angles into arcs, add the arcs, and turn back.
- When two chords, or a tangent and a secant, pass through one point, write Power of a Point.
Worked example
Point lies outside a circle with center and radius . The two tangent segments from touch the circle at and , and . What is the area of the region bounded by , , and the minor arc ?
(A) (B) (C) (D) (E)
Solution. Step 1. Each tangent is perpendicular to its radius, so .
Step 2. Right triangles and share the hypotenuse and have , so they are congruent, and . Triangle is a -- triangle with short leg , so and .
Step 3. The kite is two copies of triangle , so its area is .
Step 4. The central angle is , so sector has area .
Step 5. The region is the kite minus the sector: .
The answer is .
Pitfalls
- Putting the diameter into instead of the radius. The area comes out times too big, and the answer choices often differ by that factor of .
- Finding a sector's area with the inscribed angle instead of the central angle (which is twice as big).
- In Power of a Point, using instead of . Both lengths start at . Example: with and , the product is , not .
Traps that recur
- Assuming the inner corners lie on a circle of radius 3 (the table radius minus the mat width), rather than at distance x from the center. (2008 AMC 10A #25)
- Assuming the inner circle is the incircle of the triangle of centres, whose inradius is 1, or reading the picture's apparent symmetry as meaning the inner centre is equidistant from the three outer centres. (2025 AMC 10A #22)
- Losing a minus sign when writing the diameter, giving (sqrt 2 - 1)(sqrt 3 + 1) and the sum 5 in (B). (2025 AMC 10B #20)
- Finding only the small pipe wedged between the two and answering 1/9, forgetting that a pipe on the far side of the small one is also in contact with both. (2024 AMC 10B #21)
Problems, easiest first
Open any of them to try it here; the worked solution is on the problem's own page.