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Circles

What this skill is

Circle questions ask for circumference and area, the length of an arc or the area of a sector, the measures of angles in circles (central and inscribed), and the equation of a circle in the xyxy-plane. Angles may be given in degrees or in radians.

Key ideas

  • Everything about a circle comes from its radius. If you are given a diameter, area, or circumference, find rr first.
  • An arc or sector is a fraction of the whole circle: the central angle over 360∘360^\circ (or over 2π2\pi radians).
  • A central angle equals the measure of its arc. An inscribed angle (vertex on the circle) is half its intercepted arc.
  • A radius drawn to a point of tangency is perpendicular to the tangent line.
  • The equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 describes a circle with center (h,k)(h, k) and radius rr. If the equation is expanded, complete the square to get back to this form.

Formulas and rules

C=2πrA=πr2(x−h)2+(y−k)2=r2C = 2\pi r \qquad A = \pi r^2 \qquad (x - h)^2 + (y - k)^2 = r^2

MeasureDegrees (θ\theta)Radians (θ\theta)
arc lengthθ360⋅2πr\frac{\theta}{360} \cdot 2\pi rrθr\theta
sector areaθ360⋅πr2\frac{\theta}{360} \cdot \pi r^212r2θ\frac{1}{2} r^2 \theta
  • Inscribed angle =12×= \frac{1}{2} \times intercepted arc.
  • Completing the square: x2+10x=(x+5)2−25x^2 + 10x = (x + 5)^2 - 25. Take half the coefficient, square it.

Worked example 1 (easy)

A circle has radius 99. Find the length of the arc and the area of the sector cut off by an 80∘80^\circ central angle.

  1. Fraction of the circle: 80360=29\frac{80}{360} = \frac{2}{9}.
  2. Arc length: 29⋅2π(9)=29⋅18π=4π\frac{2}{9} \cdot 2\pi(9) = \frac{2}{9} \cdot 18\pi = 4\pi.
  3. Sector area: 29⋅π(9)2=29⋅81π=18π\frac{2}{9} \cdot \pi(9)^2 = \frac{2}{9} \cdot 81\pi = 18\pi.

Check with radians: 80∘=4π980^\circ = \frac{4\pi}{9}, and rθ=9⋅4π9=4πr\theta = 9 \cdot \frac{4\pi}{9} = 4\pi. Correct.

Worked example 2 (SAT-level)

A circle in the xyxy-plane has equation x2+y2+10x−4y=7x^2 + y^2 + 10x - 4y = 7. What are its center and radius?

  1. Group the terms: (x2+10x)+(y2−4y)=7(x^2 + 10x) + (y^2 - 4y) = 7.
  2. Complete each square: x2+10x=(x+5)2−25x^2 + 10x = (x + 5)^2 - 25 and y2−4y=(y−2)2−4y^2 - 4y = (y - 2)^2 - 4.
  3. Substitute: (x+5)2−25+(y−2)2−4=7(x + 5)^2 - 25 + (y - 2)^2 - 4 = 7, so (x+5)2+(y−2)2=36(x + 5)^2 + (y - 2)^2 = 36.
  4. The center is (−5,2)(-5, 2) and the radius is 36=6\sqrt{36} = 6.

Check: the point (1,2)(1, 2) is 66 units right of the center, so it should be on the circle: 1+4+10−8=71 + 4 + 10 - 8 = 7. Correct.

Common traps

  • Forgetting the square root. In (x−3)2+(y+1)2=64(x - 3)^2 + (y + 1)^2 = 64, the radius is 88, not 6464.
  • Center sign errors. (x+5)2(x + 5)^2 means h=−5h = -5. The center has the opposite signs of the numbers in the parentheses.
  • Not adding the completed squares to both sides. When you add 2525 and 44 on the left, add them on the right too: 7+25+4=367 + 25 + 4 = 36.
  • Using the diameter in πr2\pi r^2. A diameter of 1414 gives an area of 49π49\pi.
  • Treating an inscribed angle like a central angle. An inscribed angle intercepting a 140∘140^\circ arc measures 70∘70^\circ.
  • Mixing units. Use θ360\frac{\theta}{360} with degrees and rθr\theta with radians, never a mix.
Practice questions