Math Core

Circle problems on MATHCOUNTS and the AMC 8 mix a few formulas with a few key facts about angles, tangents and diameters. Most "shaded region" questions are just one area minus another, and the answer is usually left in terms of $\pi$ .

Circumference, area, arcs and sectors

For a circle with radius $r$ (and diameter $d = 2r$ ):

Circumference: $C = 2\pi r = \pi d$ .
Area: $A = \pi r^2$ .

An arc or sector with central angle $\theta$ is the fraction $\dfrac{\theta}{360^\circ}$ of the whole circle. So:

$\text{arc length} = \frac{\theta}{360^\circ} \cdot 2\pi r, \qquad \text{sector area} = \frac{\theta}{360^\circ} \cdot \pi r^2.$

Because area uses $r^2$ , doubling the radius multiplies the area by $4$ , and tripling it multiplies the area by $9$ . The circumference only doubles or triples.

Three facts about lines and circles

Key circle facts

Inscribed angles: an angle with its vertex on the circle is half the central angle that cuts off the same arc.
Angle in a semicircle: if one side of an inscribed triangle is a diameter, the angle across from it is a right angle.
Tangents: a tangent line is perpendicular to the radius at the point of tangency. Two tangent segments from the same outside point are equal in length.

The tangent fact creates right triangles, so it often leads straight to the Pythagorean theorem. The semicircle fact is a special case of the inscribed angle fact: a diameter cuts off a $180^\circ$ arc, and half of $180^\circ$ is $90^\circ$ .

Worked example: A sector

A sector of a circle has radius $6$ and central angle $120^\circ$ . Find its area and its perimeter.

A 120° sector of a circle with radius 6.

The sector is $\dfrac{120}{360} = \dfrac{1}{3}$ of the circle. Its area is $\dfrac{1}{3} \cdot 36\pi = 12\pi$ . Its arc is $\dfrac{1}{3} \cdot 12\pi = 4\pi$ . The perimeter includes the two radii: $6 + 6 + 4\pi = 12 + 4\pi$ .

Worked example: Circle minus square

A square is inscribed in a circle of radius $5$ . Find the area inside the circle but outside the square.

A square inscribed in a circle. Its diagonal is a diameter.

The square's diagonal is a diameter, $10$ . A square's area is half the product of its diagonals: $\dfrac{1}{2} \cdot 10 \cdot 10 = 50$ . The circle's area is $25\pi$ , so the region has area $25\pi - 50$ .

Worked example: A tangent segment

Point $P$ is $13$ units from the center $O$ of a circle with radius $5$ . A line through $P$ is tangent to the circle at $T$ . Find $PT$ .

The radius OT is perpendicular to the tangent PT.

Radius $OT$ is perpendicular to the tangent, so triangle $OTP$ has a right angle at $T$ and hypotenuse $OP = 13$ . Then $PT = \sqrt{13^2 - 5^2} = 12$ .

Worked example: A triangle on a diameter

$AB$ is a diameter of a circle with radius $5$ . Point $C$ is on the circle with $AC = 6$ . Find the area of triangle $ABC$ .

Triangle ABC with side AB along a diameter.

Since $AB$ is a diameter, $\angle ACB = 90^\circ$ . The hypotenuse is $AB = 10$ , so $BC = \sqrt{100 - 36} = 8$ . The area is $\dfrac{1}{2} \cdot 6 \cdot 8 = 24$ .

Common mistake

Check whether the problem gives the radius or the diameter. A circle with diameter $10$ has area $25\pi$ , not $100\pi$ . Circle a word like "diameter" when you read it.