Hard GRE Circle Questions Practice Questions

Concept Explanation

Hard GRE circle questions evaluate your ability to integrate properties of circles—such as area, circumference, arc length, and sector area—with complex geometric concepts like inscribed polygons and coordinate geometry. To solve these advanced problems, you must be comfortable with the relationship between the radius $r$ and other dimensions: the circumference is given by $C = 2\pi r$, and the area is $A = \pi r^2$. Beyond these basics, high-level GRE preparation requires understanding that the ratio of an arc length to the total circumference is equal to the ratio of the central angle to $360^\circ$. Similarly, the ratio of a sector's area to the total area of the circle follows this same proportionality. On the actual GRE Prep journey, you will often see circles combined with right triangles (especially those involving the Pythagorean theorem) or coordinate planes where the equation of a circle is defined as $(x - h)^2 + (y - k)^2 = r^2$, centered at $(h, k)$. For more foundational practice before tackling these difficult variations, you might find Free GRE Practice Questions helpful for refreshing core arithmetic and geometry rules.

Solved Examples

1. Example 1: Inscribed Square. A square is inscribed in a circle with an area of $36\pi$. What is the area of the square?
   1. First, find the radius of the circle using the area formula: $\pi r^2 = 36\pi$. Dividing by $\pi$ gives $r^2 = 36$, so $r = 6$.
   2. In an inscribed square, the diagonal of the square is equal to the diameter of the circle. The diameter is $2r = 12$.
   3. Let the side of the square be $s$. Using the Pythagorean theorem for the right triangle formed by two sides and the diagonal: $s^2 + s^2 = 12^2$, which simplifies to $2s^2 = 144$.
   4. Solving for the area of the square ($s^2$): $s^2 = 72$. The area is 72.
2. Example 2: Arc Length and Central Angles. In circle $O$, the length of arc $AB$ is $4\pi$ and the area of the circle is $64\pi$. What is the measure of the central angle $\angle AOB$ in degrees?
   1. Determine the radius from the area: $\pi r^2 = 64\pi$, so $r^2 = 64$ and $r = 8$.
   2. Calculate the total circumference: $C = 2\pi(8) = 16\pi$.
   3. Set up the ratio: $\ \frac{\ \text{Arc Length}}{\ \text{Circumference}} = \ \frac{\ \text{Central Angle}}{360}$.
   4. Substitute the values: $\ \frac{4\pi}{16\pi} = \ \frac{x}{360}$. This simplifies to $\ \frac{1}{4} = \ \frac{x}{360}$.
   5. Solve for $x$: $x = 90^\circ$.
3. Example 3: Tangent Lines and Right Triangles. A line is tangent to circle $C$ at point $P$. The center of the circle is at $(0, 0)$ and the radius is 5. If a point $Q$ on the tangent line is 13 units from the center of the circle, what is the distance from $P$ to $Q$?
   1. Recognize that a radius drawn to a point of tangency is perpendicular to the tangent line. This forms a right triangle $\ riangle CPQ$.
   2. The hypotenuse is the distance from the center to point $Q$, which is 13.
   3. One leg is the radius $CP = 5$. The other leg is the distance $PQ$.
   4. Apply the Pythagorean theorem: $5^2 + PQ^2 = 13^2$.
   5. $25 + PQ^2 = 169$, so $PQ^2 = 144$ and $PQ = 12$.

Practice Questions

Test your skills with these hard GRE circle questions. Reviewing GRE Practice Questions with Explanations can assist in identifying common traps in these geometry problems.

1. A circle is tangent to the x-axis at $(4, 0)$ and tangent to the y-axis at $(0, 4)$. What is the area of the region in the first quadrant that is outside the circle but inside the square formed by the origin and the tangent points?
2. If the circumference of Circle A is 20% greater than the circumference of Circle B, by what percent is the area of Circle A greater than the area of Circle B?
3. A chord of length 16 is 6 units away from the center of a circle. What is the circumference of this circle?

4. An equilateral triangle is inscribed in a circle with radius 10. What is the area of the triangle?
5. In a coordinate plane, a circle has the equation $x^2 + y^2 - 4x + 6y = 3$. What is the radius of this circle?
6. A circular track has a radius of 50 meters. If a runner completes an arc of 120 degrees, what is the distance covered by the runner?
7. Quantity A: The area of a circle with diameter $d$. Quantity B: The area of a square with diagonal $d$. Which quantity is greater?
8. A circle is inscribed inside a right triangle with side lengths 6, 8, and 10. What is the radius of the inscribed circle?
9. Two circles, $C1$ and $C2$, have radii of 3 and 8 respectively. If the distance between their centers is 13, what is the length of their common external tangent?
10. The area of a sector of a circle with radius 12 is $24\pi$. What is the perimeter of this sector?

Answers & Explanations

1. Answer: $16 - 4\pi$. The square formed by the origin $(0,0)$, $(4,0)$, $(4,4)$, and $(0,4)$ has an area of $4 \ \times 4 = 16$. The circle tangent to these points has a radius of 4 and its center at $(4,4)$ or $(0,0)$ etc., but in the first quadrant, the quarter-circle inside this square has an area of $\ \frac{1}{4} \pi (4^2) = 4\pi$. Subtracting the quarter-circle from the square gives $16 - 4\pi$.
2. Answer: 44%. Let the radius of Circle B be $r$. Its circumference is $2\pi r$. Circle A's circumference is $1.2(2\pi r)$, so its radius is $1.2r$. The area of Circle B is $\pi r^2$. The area of Circle A is $\pi (1.2r)^2 = 1.44 \pi r^2$. This is a 44% increase.
3. Answer: $20\pi$. A line from the center perpendicular to a chord bisects the chord. This forms a right triangle with one leg 6 (distance from center) and another leg 8 (half the chord). The hypotenuse is the radius: $6^2 + 8^2 = r^2 \ \rightarrow 36 + 64 = 100 \ \rightarrow r = 10$. Circumference is $2\pi(10) = 20\pi$.
4. Answer: $75\sqrt{3}$. For an equilateral triangle in a circle, the radius $R = \ \frac{s}{\sqrt{3}}$, where $s$ is the side. Thus, $10 = \ \frac{s}{\sqrt{3}} \ \rightarrow s = 10\sqrt{3}$. The area of an equilateral triangle is $\ \frac{s^2\sqrt{3}}{4} = \ \frac{(10\sqrt{3})^2\sqrt{3}}{4} = \ \frac{300\sqrt{3}}{4} = 75\sqrt{3}$.
5. Answer: 4. Complete the square: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 3 + 4 + 9$. This becomes $(x-2)^2 + (y+3)^2 = 16$. The radius is $\sqrt{16} = 4$.
6. Answer: $\ \frac{100\pi}{3}$. Distance is arc length: $L = \ \frac{120}{360} \ \times 2\pi(50) = \ \frac{1}{3} \ \times 100\pi = \ \frac{100\pi}{3}$.
7. Answer: Quantity A. Area of circle with diameter $d$ is $\pi (d/2)^2 = \ \frac{\pi d^2}{4} \approx 0.785 d^2$. Area of square with diagonal $d$ is $\ \frac{d^2}{2} = 0.5 d^2$. Since $0.785 > 0.5$, Quantity A is greater.
8. Answer: 2. For a right triangle, the inradius $r = \ \frac{a + b - c}{2}$. Here $a=6, b=8, c=10$. So $r = \ \frac{6+8-10}{2} = \ \frac{4}{2} = 2$.
9. Answer: 12. The formula for the length of an external tangent is $𝑑 = \sqrt{D^2 - (R - r)^2}$. Here, $D = 13, R = 8, r = 3$. So $𝑑 = \sqrt{13^2 - (8-3)^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.
10. Answer: $4\pi + 24$. Area of sector $24\pi = \ \frac{\ heta}{360} \pi (12^2) \ \rightarrow 24\pi = \ \frac{\ heta}{360} 144\pi$. Thus $\ \frac{\ heta}{360} = \ \frac{24}{144} = \ \frac{1}{6}$. Arc length is $\ \frac{1}{6} (2\pi \ \times 12) = 4\pi$. Perimeter includes two radii: $4\pi + 12 + 12 = 4\pi + 24$.

1. A circle is inscribed in a square with side length 10. What is the area of the circle?

• A\( 100\pi \)
• B\( 50\pi \)
• C\( 25\pi \)
• D\( 10\pi \)

Frequently Asked Questions

How do I find the area of a shaded region involving a circle and a square?

Calculate the area of both shapes individually using the given dimensions and then subtract the smaller area from the larger one. Ensure you identify whether the circle is inscribed (diameter equals side) or circumscribed (diameter equals diagonal).

What is the relationship between central angles and arc lengths on the GRE?

The arc length is a fraction of the total circumference, determined by the ratio of the central angle to 360 degrees. This linear relationship also applies to the area of a sector relative to the total area of the circle.

Can I use the Pythagorean theorem for circle problems?

Yes, the Pythagorean theorem is frequently used when a radius meets a tangent line at a right angle or when finding the radius using a chord and its distance from the center. These scenarios create right triangles where the radius often serves as the hypotenuse or a leg.

What is the equation of a circle in the coordinate plane?

The standard equation is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ represents the center and $r$ is the radius. You may need to complete the square if the equation is provided in an expanded polynomial format.

How does the GRE test inscribed angles versus central angles?

An inscribed angle has its vertex on the circle's edge, while a central angle has its vertex at the center. An inscribed angle is always half the measure of the central angle that intercepts the same arc.