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    Hard GRE Circle Questions Practice Questions

    July 8, 202611 min read52 views
    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 rr and other dimensions: the circumference is given by C=2πrC = 2\pi r, and the area is A=πr2A = \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 360360^\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 (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, centered at (h,k)(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 36π36\pi. What is the area of the square?
      1. First, find the radius of the circle using the area formula: πr2=36π\pi r^2 = 36\pi. Dividing by π\pi gives r2=36r^2 = 36, so r=6r = 6.
      2. In an inscribed square, the diagonal of the square is equal to the diameter of the circle. The diameter is 2r=122r = 12.
      3. Let the side of the square be ss. Using the Pythagorean theorem for the right triangle formed by two sides and the diagonal: s2+s2=122s^2 + s^2 = 12^2, which simplifies to 2s2=1442s^2 = 144.
      4. Solving for the area of the square (s2s^2): s2=72s^2 = 72. The area is 72.
    2. Example 2: Arc Length and Central Angles. In circle OO, the length of arc ABAB is 4π4\pi and the area of the circle is 64π64\pi. What is the measure of the central angle AOB\angle AOB in degrees?
      1. Determine the radius from the area: πr2=64π\pi r^2 = 64\pi, so r2=64r^2 = 64 and r=8r = 8.
      2. Calculate the total circumference: C=2π(8)=16πC = 2\pi(8) = 16\pi.
      3. Set up the ratio:   Arc Length Circumference=  Central Angle360\ \frac{\ \text{Arc Length}}{\ \text{Circumference}} = \ \frac{\ \text{Central Angle}}{360}.
      4. Substitute the values:  4π16π= x360\ \frac{4\pi}{16\pi} = \ \frac{x}{360}. This simplifies to  14= x360\ \frac{1}{4} = \ \frac{x}{360}.
      5. Solve for xx: x=90x = 90^\circ.
    3. Example 3: Tangent Lines and Right Triangles. A line is tangent to circle CC at point PP. The center of the circle is at (0,0)(0, 0) and the radius is 5. If a point QQ on the tangent line is 13 units from the center of the circle, what is the distance from PP to QQ?
      1. Recognize that a radius drawn to a point of tangency is perpendicular to the tangent line. This forms a right triangle  riangleCPQ\ riangle CPQ.
      2. The hypotenuse is the distance from the center to point QQ, which is 13.
      3. One leg is the radius CP=5CP = 5. The other leg is the distance PQPQ.
      4. Apply the Pythagorean theorem: 52+PQ2=1325^2 + PQ^2 = 13^2.
      5. 25+PQ2=16925 + PQ^2 = 169, so PQ2=144PQ^2 = 144 and PQ=12PQ = 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)(4, 0) and tangent to the y-axis at (0,4)(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?

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    Practice GRE Questions
    1. An equilateral triangle is inscribed in a circle with radius 10. What is the area of the triangle?
    2. In a coordinate plane, a circle has the equation x2+y24x+6y=3x^2 + y^2 - 4x + 6y = 3. What is the radius of this circle?
    3. 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?
    4. Quantity A: The area of a circle with diameter dd. Quantity B: The area of a square with diagonal dd. Which quantity is greater?
    5. A circle is inscribed inside a right triangle with side lengths 6, 8, and 10. What is the radius of the inscribed circle?
    6. Two circles, C1C1 and C2C2, have radii of 3 and 8 respectively. If the distance between their centers is 13, what is the length of their common external tangent?
    7. The area of a sector of a circle with radius 12 is 24π24\pi. What is the perimeter of this sector?

    Answers & Explanations

    1. Answer: 164π16 - 4\pi. The square formed by the origin (0,0)(0,0), (4,0)(4,0), (4,4)(4,4), and (0,4)(0,4) has an area of 4 ×4=164 \ \times 4 = 16. The circle tangent to these points has a radius of 4 and its center at (4,4)(4,4) or (0,0)(0,0) etc., but in the first quadrant, the quarter-circle inside this square has an area of  14π(42)=4π\ \frac{1}{4} \pi (4^2) = 4\pi. Subtracting the quarter-circle from the square gives 164π16 - 4\pi.
    2. Answer: 44%. Let the radius of Circle B be rr. Its circumference is 2πr2\pi r. Circle A's circumference is 1.2(2πr)1.2(2\pi r), so its radius is 1.2r1.2r. The area of Circle B is πr2\pi r^2. The area of Circle A is π(1.2r)2=1.44πr2\pi (1.2r)^2 = 1.44 \pi r^2. This is a 44% increase.
    3. Answer: 20π20\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: 62+82=r2 36+64=100 r=106^2 + 8^2 = r^2 \ \rightarrow 36 + 64 = 100 \ \rightarrow r = 10. Circumference is 2π(10)=20π2\pi(10) = 20\pi.
    4. Answer: 75375\sqrt{3}. For an equilateral triangle in a circle, the radius R= s3R = \ \frac{s}{\sqrt{3}}, where ss is the side. Thus, 10= s3 s=10310 = \ \frac{s}{\sqrt{3}} \ \rightarrow s = 10\sqrt{3}. The area of an equilateral triangle is  s234= (103)234= 30034=753\ \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: (x24x+4)+(y2+6y+9)=3+4+9(x^2 - 4x + 4) + (y^2 + 6y + 9) = 3 + 4 + 9. This becomes (x2)2+(y+3)2=16(x-2)^2 + (y+3)^2 = 16. The radius is 16=4\sqrt{16} = 4.
    6. Answer:  100π3\ \frac{100\pi}{3}. Distance is arc length: L= 120360 ×2π(50)= 13 ×100π= 100π3L = \ \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 dd is π(d/2)2= πd240.785d2\pi (d/2)^2 = \ \frac{\pi d^2}{4} \approx 0.785 d^2. Area of square with diagonal dd is  d22=0.5d2\ \frac{d^2}{2} = 0.5 d^2. Since 0.785>0.50.785 > 0.5, Quantity A is greater.
    8. Answer: 2. For a right triangle, the inradius r= a+bc2r = \ \frac{a + b - c}{2}. Here a=6,b=8,c=10a=6, b=8, c=10. So r= 6+8102= 42=2r = \ \frac{6+8-10}{2} = \ \frac{4}{2} = 2.
    9. Answer: 12. The formula for the length of an external tangent is 𝑑=D2(Rr)2𝑑 = \sqrt{D^2 - (R - r)^2}. Here, D=13,R=8,r=3D = 13, R = 8, r = 3. So 𝑑=132(83)2=16925=144=12𝑑 = \sqrt{13^2 - (8-3)^2} = \sqrt{169 - 25} = \sqrt{144} = 12.
    10. Answer: 4π+244\pi + 24. Area of sector 24π=  heta360π(122) 24π=  heta360144π24\pi = \ \frac{\ heta}{360} \pi (12^2) \ \rightarrow 24\pi = \ \frac{\ heta}{360} 144\pi. Thus   heta360= 24144= 16\ \frac{\ heta}{360} = \ \frac{24}{144} = \ \frac{1}{6}. Arc length is  16(2π ×12)=4π\ \frac{1}{6} (2\pi \ \times 12) = 4\pi. Perimeter includes two radii: 4π+12+12=4π+244\pi + 12 + 12 = 4\pi + 24.
    Interactive quizQuestion 1 of 5

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

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    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 (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) represents the center and rr 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.

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    Practice GRE Questions

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