# Circles - GMAT Math Study Guide

## Definitions

• Circle - An infinite set of points that are all equidistant from a point called the center.
• Circumference - The edge (or boundary) of a circle.
• Radius - A line from the center of a circle to the edge of the circle.
• Diameter - A line that passes through the center of the circle and has its endpoints at the edge of the circle.
• Chord - A line that lies within the circle and connects two points on the edge of the circle.
• Central Angle - An angle whose vertex is the center of the circle.
• Arc - A portion of the circumference of a circle (i.e., a segment of the edge of a circle).
• Sector of a Circle - A slice (or sliver) of the circle enclosed on one side by the edge of the circle and on the other sides by radii.
• Center of a Circle - The point in a circle from which all points on the edge of the circle are equidistant.
• Circumscribed Circle - A circle that encompasses (or surrounds from the outside) a polygon such that the circle's circumference intersects the vertices of the polygon.
• Inscribed Circle - An inscribed circle is a circle that lies inside a figure such that points on the edge of the circle are tangent to the sides of the figure.
• Concentric Circles - Circles with the same center point but not necessarily the same radius length.

## Properties of a Circle

A circle is formed by an infinite number of points that are equidistant from a center. The length of the equidistant points from the center is the radius, r. ### Formulas for Circles

Area = πr2
Diameter, d = 2r
Circumference = 2πr = πd
Central Angle = 2(Inscribed Angle)
Area of Sector = (x/360)πr2
Length of an Arc = (x/360)Circumference = (x/360)2πr

## Arcs & Sectors

### Arcs

An arc is a portion of the edge of a circle. For example, the portion of the circle between A and B is called arc AB. The length of the arc is proportional to the central angle that forms the arc (i.e., angle ACB in this example). In the above diagram, if radius CB is 9cm and angle ACB is 60°, what is the length of arc AB?

1. Find the circumference
C = 2π(9) = 18π
2. Plug into the length of an arc formula
(60/360)(18π)
(1/6)(18π)
3π

### Sectors

A sector is a portion of the area of a circle. For example, the portion of the circle between center C and arc AB forms a sector (this sector is shaded in black below). The area of the sector is proportional to the central angle that forms it. In the figure above, if radius CA is 9cm and angle ACB is 60°, what is the area of the region shaded in black?

1. Find the area
A = π(9)2 = 81π
2. Plug into the area of a sector formula
(60/360)81π
(1/6)(81π)
(81/6)π

## Inscribed Angle & Central Angle

A central angle is an angle whose vertex is the center of the circle and whose endpoints are the edge of the circle. Angle ACB is a central angle. An inscribed angle is an angle whose vertex lies on the edge of the circle and whose endpoints lie on another part of the edge of the circle. Angle ADB and angle AEB are both examples of inscribed angles. For a central angle and an inscribed angle with the same endpoints:
1. All inscribed angles with the same endpoints are equal
2. Inscribed Angle = (1/2)(Central Angle)

### Inscribed Triangle

As a result of the equality mentioned above between an inscribed angle and half of the measurement of a central angle, the following property holds true: if a triangle is inscribed in a circle such that one side of that triangle is a diameter of the circle, then the angle of the triangle that is opposite the diameter is a right angle. The reason that angle ADB is a right angle is because central angle ACB is 180° and ADB is an inscribed angle whose endpoints are the same as ACB.

Consequently:

For the above to hold true: (1) C must be the center of the circle (2) AB must be a diameter of the center

## Inscribed Circles & Circumscribed Circles

### Inscribed Circle

An inscribed circle is a circle that lies inside a figure such that points on the edge of the circle are tangent to the sides of the figure. For example, the following is a circle inscribed in a square. ### Circumscribed Circles

A circumscribed circle is a circle that encompasses a polygon such that the circle touches all the vertices of the polygon. The following is a circle circumscribed around a rectangle. ## Types of GMAT Problems

1. Polygons Inscribed in Circles

A shape is said to be inscribed in a circle if each vertex of the shape lies on the circle. If a triangle is inscribed inside of a circle, and the base of the triangle is also a diameter of the circle, then the triangle is a right triangle.

Find the area of the black region. Assume that the base of the triangle is a diameter of the circle and the radius of the circle is 12.5 A) 121π-160 B) 156.25π-150 C) 120.25π-145 D) 152.75π-132 E) 133.5π-125
1. In order to determine the area of the shaded region (As), the area of the circle (Ac) and the area of the triangle (At) must be discovered.
Ac = π*r2 = π*12.52 = 156.25π
At = .5*b*h, where b is the base and h is the height.
2. The base of the triangle is the diameter of the circle, which is 2*r = 2*12.5 = 25.
The height of the triangle (y) is the length of the line which is perpendicular to the base and goes through the opposite vertex.
3. The height of the triangle is part of a right triangle.
162 + y2 = x2
However, to determine the height, y, the other two sides of the right triangle must be known. Yet we do not know x, so we must solve for x before we can solve for y.
4. Because the larger triangle with sides 15, x, and 25 has a base as the diameter of the circle, it is a right triangle and the angle opposite the diameter must be 90. Thus, the Pythagorean theorem can be used to find the length of x.
x2 + 152 = 252
Rather than do the calculations, notice that the triangle is a 3-4-5 triangle (multiplied by 5). Thus, x must be 20 (=4*5).
5. Two sides of the right triangle containing y are known and the Pythagorean theorem can be used once more.
162 + y2 = x2
162 + y2 = 202
Notice once again that this is a 3-4-5 triangle (multiplied by 4). Thus, y must be 12 (=3*4).
6. With the base and height calculated, the area of the triangle can be calculated and subtracted from the area of the circle.
At = .5 * 25 * 12 = 150
As = 156.25π-150