4.3 Circles & Inscribed Figures

Key Takeaways

  • Circle circumference is C = 2πr = πd and area is A = πr²; doubling the radius quadruples the area while doubling the circumference.
  • Arc length is given by (θ/360°) × 2πr and sector area is given by (θ/360°) × πr², where θ is the central angle in degrees.
  • A tangent line to a circle is always perpendicular (90°) to the radius drawn to the point of tangency.
  • An inscribed angle is equal to half the measure of its intercepted central angle, and any angle inscribed in a semicircle is a right angle (90°).
  • For a square inscribed in a circle, the square's diagonal equals the circle's diameter (d = s√2 = 2r); for a circle inscribed in a square, the circle's diameter equals the square's side length (2r = s).
Last updated: July 2026

4.3 Circles & Inscribed Figures

Circle geometry is a staple of the GRE Quantitative Reasoning section. Circles often appear alongside inscribed or circumscribed polygons, shaded region problems, and arc/sector calculations. Mastering the geometric relationships between radius, diameter, circumference, area, central angles, inscribed angles, tangent lines, and inscribed polygons will enable you to solve complex circle problems with speed and accuracy.


1. Core Circle Definitions & Fundamental Formulas

A circle is the set of all points in a two-dimensional plane that are equidistant from a fixed point called the center.

  • Radius ($r$): The distance from the center of the circle to any point on its circumference.
  • Diameter ($d$): A straight line segment passing through the center of the circle with both endpoints on the circle. d=2rorr=d2d = 2r \quad \text{or} \quad r = \frac{d}{2}
  • Circumference ($C$): The perimeter or linear distance around the outside of the circle. C=2πr=πdC = 2\pi r = \pi d
  • Area ($A$): The total two-dimensional space enclosed within the circle. A=πr2A = \pi r^2

Proportional Relationships & Scaling

Understanding how changing the radius scales circumference and area is a frequent GRE shortcut:

  • If the radius $r$ is multiplied by a factor of $k$:
    • The diameter $d$ is multiplied by $k$.
    • The circumference $C$ is multiplied by $k$.
    • The area $A$ is multiplied by $k^2$. Example: Tripling the radius ($k = 3$) triples the circumference ($3\times$) and multiplies the area by $3^2 = 9$ ($9\times$).

2. Lines, Chords, Secants, and Tangents

  1. Chord: A line segment whose endpoints both lie on the circle. The diameter is the longest possible chord in any circle.
  2. Secant: A line that intersects a circle at two distinct points.
  3. Tangent: A line that intersects a circle at exactly one point, known as the point of tangency.

Tangent Radius Theorem: A line tangent to a circle is always perpendicular ($90^\circ$) to the radius drawn to the point of tangency.

GRE Application: When a tangent line appears in a diagram, draw a radius to the point of tangency immediately. This creates a $90^\circ$ right angle, allowing you to use the Pythagorean Theorem or special right triangle rules!


3. Central Angles vs. Inscribed Angles

  • Central Angle: An angle whose vertex is at the center of the circle and whose sides are radii. The degree measure of a central angle equals the degree measure of the arc it intercepts. Central Angle θ=Arc Measure\text{Central Angle } \theta = \text{Arc Measure}
  • Inscribed Angle: An angle whose vertex lies on the circle and whose sides are chords.

Inscribed Angle Theorem: An inscribed angle is equal to exactly half the measure of the central angle that subtends (intercepts) the same arc.

Inscribed Angle=12×Central Angle=12×Intercepted Arc Measure\text{Inscribed Angle} = \frac{1}{2} \times \text{Central Angle} = \frac{1}{2} \times \text{Intercepted Arc Measure}

Thales's Theorem (Inscribed Semicircle Theorem)

Any angle inscribed in a semicircle (an inscribed angle that intercepts a diameter) is always a right angle ($90^\circ$).

GRE Application: If a triangle is inscribed in a circle such that one of its sides is the diameter of the circle, the triangle is automatically a right triangle with the diameter as its hypotenuse.


4. Arc Length & Sector Area

An arc is a continuous portion of the circumference of a circle. A sector is a pie-shaped region bounded by two radii and an arc.

For a central angle of $\theta$ degrees:

Arc Length Formula

Arc Length=(θ360)×2πr\text{Arc Length} = \left(\frac{\theta}{360^\circ}\right) \times 2\pi r

Sector Area Formula

Sector Area=(θ360)×πr2\text{Sector Area} = \left(\frac{\theta}{360^\circ}\right) \times \pi r^2

Summary Table: Arcs and Sectors

Angle Fraction ($\frac{\theta}{360^\circ}$)Central Angle ($\theta$)Arc LengthSector Area
$\frac{1}{4}$ (Quarter Circle)$90^\circ$$\frac{1}{4}(2\pi r) = \frac{\pi r}{2}$$\frac{1}{4}\pi r^2$
$\frac{1}{3}$$120^\circ$$\frac{1}{3}(2\pi r) = \frac{2\pi r}{3}$$\frac{1}{3}\pi r^2$
$\frac{1}{2}$ (Semicircle)$180^\circ$$\pi r$$\frac{1}{2}\pi r^2$
$\frac{\theta}{360^\circ}$$\theta^\circ$$\frac{\theta}{360^\circ} \cdot 2\pi r$$\frac{\theta}{360^\circ} \cdot \pi r^2$

5. Inscribed and Circumscribed Figures

Combining circles with polygons creates powerful geometric relationships:

1. Square Inscribed in a Circle (Circle Circumscribed Around Square)

  • All four vertices of the square lie on the circle.
  • Key Geometric Relationship: The diagonal of the square is equal to the diameter of the circle. Diagonal d=s2=2r\text{Diagonal } d = s\sqrt{2} = 2r Square side s=r2,Circle Radius r=s22\text{Square side } s = r\sqrt{2}, \quad \text{Circle Radius } r = \frac{s\sqrt{2}}{2}

2. Circle Inscribed in a Square (Square Circumscribed Around Circle)

  • The circle is tangent to all four sides of the square.
  • Key Geometric Relationship: The side length of the square is equal to the diameter of the circle. s=2r    r=s2s = 2r \quad \implies \quad r = \frac{s}{2} Area of Square=s2=(2r)2=4r2\text{Area of Square} = s^2 = (2r)^2 = 4r^2 Area of Circle=πr2\text{Area of Circle} = \pi r^2 Ratio of Circle Area to Square Area=πr24r2=π478.5%\text{Ratio of Circle Area to Square Area} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4} \approx 78.5\%

6. Worked GRE Step-by-Step Examples

Example 1: Shaded Region (Sector minus Triangle)

Problem: A circle has radius $r = 6$. A central angle of $90^\circ$ creates a sector $OAB$, where $O$ is the center of the circle and $A, B$ are points on the circle. Find the exact area of the segment bounded by chord $AB$ and arc $AB$ (the shaded region outside triangle $OAB$).

Solution:

  1. Calculate Sector Area: Sector Area=(90360)×π(62)=14×36π=9π\text{Sector Area} = \left(\frac{90^\circ}{360^\circ}\right) \times \pi (6^2) = \frac{1}{4} \times 36\pi = 9\pi
  2. Calculate Triangle Area:
    • Triangle $OAB$ is a right isosceles triangle with base $OA = 6$ and height $OB = 6$. Triangle Area=12×6×6=18\text{Triangle Area} = \frac{1}{2} \times 6 \times 6 = 18
  3. Subtract Triangle Area from Sector Area: Shaded Segment Area=Sector AreaTriangle Area=9π18\text{Shaded Segment Area} = \text{Sector Area} - \text{Triangle Area} = 9\pi - 18

Example 2: Square Inscribed in a Circle Ratio

Problem: A square with side length 8 is inscribed in a circle. What is the area of the region inside the circle but outside the square?

Solution:

  1. Find Square Area: Areasquare=82=64\text{Area}_{\text{square}} = 8^2 = 64
  2. Find Circle Diameter and Radius:
    • The diagonal of the inscribed square equals the diameter of the circle: d=s2=82d = s\sqrt{2} = 8\sqrt{2}
    • Therefore, radius $r = \frac{d}{2} = 4\sqrt{2}$.
  3. Calculate Circle Area: Areacircle=πr2=π(42)2=π(16×2)=32π\text{Area}_{\text{circle}} = \pi r^2 = \pi (4\sqrt{2})^2 = \pi (16 \times 2) = 32\pi
  4. Calculate Remaining Region Area: Remaining Area=AreacircleAreasquare=32π64\text{Remaining Area} = \text{Area}_{\text{circle}} - \text{Area}_{\text{square}} = 32\pi - 64

7. GRE Exam Strategies & Common Pitfalls

  • Keep Answers in Terms of $\pi$: Unless an answer option or numeric entry prompt explicitly demands a decimal approximation (e.g., using $\pi \approx 3.14$), leave $\pi$ as a symbol throughout your work.
  • Look for Right Triangles: Tangents meeting radii at $90^\circ$ and inscribed angles subtending diameters ($90^\circ$) are the two most common ways the GRE hides right triangles inside circle problems.
  • Radius is Always Uniform: Any line drawn from the center of a circle to a point on its boundary has length $r$. Drawing additional radii to key vertices often forms isosceles triangles, simplifying complex angle puzzles.
Test Your Knowledge

In a circle with center O, angle AOB is a central angle measuring 80°. Point C lies on the major arc AB. What is the measure of the inscribed angle ACB?

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Test Your Knowledge

A circle has a radius of 12. What is the area of a sector with a central angle of 60°?

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Test Your Knowledge

A square is inscribed in a circle of radius 5. What is the area of the square?

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Test Your Knowledge

A line is tangent to a circle of radius 7 at point T. Point P lies on the tangent line such that the distance from P to the center of the circle is 25. What is the distance PT from P to the point of tangency?

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