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🎯⭐ INTERACTIVE LESSON

Circles

Learn step-by-step with interactive practice!

Circles - Complete Interactive Lesson

Part 1: Circle Equations

⭕ Circle Equations

Part 1 of 7 — Standard Form, Center, Radius & Completing the Square

Circle equations appear on almost every SAT. The standard form is the key:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2
ComponentMeaning
(h,k)(h, k)Center of the circle
rrRadius
r2r^2The number on the right side

Important: The signs in (x−h)(x - h) and (y−k)(y - k) are subtractions, so if the equation has (x+3)2(x + 3)^2, that means h=−3h = -3.

Reading Center & Radius — Worked Examples

Example 1: Find the center and radius of (x−4)2+(y+1)2=25(x - 4)^2 + (y + 1)^2 = 25.

  1. Compare to (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
  2. h=4h = 4, k=−1k = -1 (note the +1+1 means k=−1k = -1).
  3. r2=25r^2 = 25, so r=5r = 5.
  4. Center: (4,−1)(4, -1), Radius: 55.

Example 2: Write the equation of a circle with center (−2,7)(-2, 7) and radius 33.

(x−(−2))2+(y−7)2=32(x - (-2))^2 + (y - 7)^2 = 3^2 (x+2)2+(y−7)2=9(x + 2)^2 + (y - 7)^2 = 9


SAT Tip: If you see r2=36r^2 = 36, the radius is 66, not 3636. The SAT loves including r2r^2 as a wrong answer for the radius.

Practice — Reading Circle Equations 🔍

Completing the Square — Key SAT Skill

Many SAT problems give the circle in general form and ask you to find the center or radius. You must complete the square.

Example: Find the center and radius of x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0.

  1. Group xx and yy terms: (x2−6x)+(y2+4y)=12(x^2 - 6x) + (y^2 + 4y) = 12.
  2. Complete the square for xx: half of −6-6 is −3-3; (−3)2=9(-3)^2 = 9.
  3. Complete the square for yy: half of 44 is 22; (2)2=4(2)^2 = 4.
  4. Add both to each side: (x2−6x+9)+(y2+4y+4)=12+9+4(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4.
  5. Factor: (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25.
  6. Center: (3,−2)(3, -2), Radius: 55.

The pattern: For x2+bxx^2 + bx, add (b2)2\left(\frac{b}{2}\right)^2 to both sides.

Complete the square practice. 🧮

Consider the circle x2+y2+8x−2y−8=0x^2 + y^2 + 8x - 2y - 8 = 0.

  1. What is the xx-coordinate of the center?

  2. What is the yy-coordinate of the center?

  3. What is the radius?

Match each equation to its center. 🔍

SAT-Style Questions 📋

Part 2: Arc Length & Sectors

📐 Arc Length & Sector Area

Part 2 of 7 — Formulas, Degree ↔ Radian Conversion

When a central angle θ\theta (in radians) intercepts an arc of a circle with radius rr:

QuantityFormula (radians)Formula (degrees)
Arc lengths=rθs = r\thetas=θ360⋅2πrs = \frac{\theta}{360} \cdot 2\pi r
Sector areaA=12r2θA = \frac{1}{2}r^2\thetaA=θ360⋅πr2A = \frac{\theta}{360} \cdot \pi r^2

Degree ↔ Radian Conversion:

radians=degrees×π180degrees=radians×180π\text{radians} = \text{degrees} \times \frac{\pi}{180} \qquad \text{degrees} = \text{radians} \times \frac{180}{\pi}
DegreesRadians
30°30°π6\frac{\pi}{6}
45°45°π4\frac{\pi}{4}
60°60°π3\frac{\pi}{3}
90°90°π2\frac{\pi}{2}
180°180°π\pi
360°360°2π2\pi

Worked Examples

Example 1 — Arc Length: A circle has radius 1010 cm. Find the length of the arc intercepted by a central angle of π3\frac{\pi}{3} radians.

s=rθ=10⋅π3=10π3≈10.47 cms = r\theta = 10 \cdot \frac{\pi}{3} = \frac{10\pi}{3} \approx 10.47 \text{ cm}

Example 2 — Sector Area: A pizza slice has radius 88 inches and a central angle of 45°45°.

  1. Convert: 45°=45×π180=π445° = 45 \times \frac{\pi}{180} = \frac{\pi}{4} radians.
  2. Area: A=12(8)2⋅π4=12(64)⋅π4=8π≈25.13A = \frac{1}{2}(8)^2 \cdot \frac{\pi}{4} = \frac{1}{2}(64)\cdot \frac{\pi}{4} = 8\pi \approx 25.13 sq in.

Example 3 — Degree Method: A sector has radius 66 and central angle 120°120°. A=120360⋅π(6)2=13⋅36π=12πA = \frac{120}{360} \cdot \pi(6)^2 = \frac{1}{3} \cdot 36\pi = 12\pi


SAT Tip: Always check whether the angle is given in degrees or radians. Using the wrong formula is the #1 mistake on these problems.

Practice — Arc Length & Sector Area 🔍

Conversion Practice — Degrees ↔ Radians

Example: Convert 150°150° to radians.

150×π180=150π180=5π6150 \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}

Example: Convert 3π4\frac{3\pi}{4} radians to degrees.

3π4×180π=3×1804=135°\frac{3\pi}{4} \times \frac{180}{\pi} = \frac{3 \times 180}{4} = 135°

Shortcut: To convert degrees → radians, divide by 180 and multiply by π\pi. To go back, divide by π\pi and multiply by 180.

Compute each value (give exact answers as integers or simplified fractions). 🧮

  1. Convert 270°270° to radians. Enter the numerator when written as ?π2\frac{?\pi}{2}.

  2. A circle has radius 44 and arc length 6π6\pi. What is the central angle in degrees?

  3. The area of a full circle with radius 1010 is 100π100\pi. What fraction of the area is a sector with central angle 72°72°? Enter as a fraction (e.g. 1/5).

Match each degree measure to its radian equivalent. 🔍

SAT-Style Questions 📋

Part 3: Right Triangle Trig

📏 Right Triangle Trigonometry

Part 3 of 7 — SOH-CAH-TOA, Finding Sides & Angles, Special Triangles

For a right triangle with an acute angle θ\theta:

RatioFormulaMnemonic
sin⁡θ\sin \thetaoppositehypotenuse\frac{\text{opposite}}{\text{hypotenuse}}SOH
cos⁡θ\cos \thetaadjacenthypotenuse\frac{\text{adjacent}}{\text{hypotenuse}}CAH
tan⁡θ\tan \thetaoppositeadjacent\frac{\text{opposite}}{\text{adjacent}}TOA

Also: tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

SAT Tip: The SAT provides the reference formulas for special triangles on the formula sheet, but memorizing them saves precious time.

Special Right Triangles

45-45-90 Triangle:

sides=1:1:2\text{sides} = 1 : 1 : \sqrt{2}

If a leg is aa, the hypotenuse is a2a\sqrt{2}.

sin⁡45°=cos⁡45°=22tan⁡45°=1\sin 45° = \cos 45° = \frac{\sqrt{2}}{2} \qquad \tan 45° = 1

30-60-90 Triangle:

sides=1:3:2\text{sides} = 1 : \sqrt{3} : 2

  • Short leg (opposite 30°30°) =x= x
  • Long leg (opposite 60°60°) =x3= x\sqrt{3}
  • Hypotenuse =2x= 2x
Anglesin⁡\sincos⁡\costan⁡\tan
30°30°12\frac{1}{2}32\frac{\sqrt{3}}{2}13\frac{1}{\sqrt{3}}
60°60°32\frac{\sqrt{3}}{2}12\frac{1}{2}3\sqrt{3}

Worked Examples

Example 1: In a right triangle, sin⁡A=513\sin A = \frac{5}{13}. Find cos⁡A\cos A.

  1. Opposite =5= 5, hypotenuse =13= 13.
  2. Adjacent =132−52=169−25=144=12= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.
  3. cos⁡A=1213\cos A = \frac{12}{13}.

Example 2: A ladder leans against a wall, making a 60°60° angle with the ground. If the foot of the ladder is 44 meters from the wall, how long is the ladder?

  1. The side adjacent to 60°60° is 44 m, and we want the hypotenuse.
  2. cos⁡60°=4hyp⇒12=4hyp⇒hyp=8\cos 60° = \frac{4}{\text{hyp}} \Rightarrow \frac{1}{2} = \frac{4}{\text{hyp}} \Rightarrow \text{hyp} = 8 m.

Example 3: A square has side length 66. What is the length of its diagonal?

The diagonal creates a 45-45-9045\text{-}45\text{-}90 triangle: d=62d = 6\sqrt{2}.

Practice — Right Triangle Trig 🔍

Compute each value. 🧮

  1. In a right triangle, the legs are 99 and 1212. What is the hypotenuse?

  2. sin⁡30°+cos⁡60°=?\sin 30° + \cos 60° = ? (enter a decimal or fraction)

  3. A 45-45-9045\text{-}45\text{-}90 triangle has hypotenuse 1010. What is the length of each leg? Round to the nearest hundredth.

Match each trig expression to its value. 🔍

SAT-Style Questions 📋

Part 4: Unit Circle Basics

🔵 Unit Circle Basics

Part 4 of 7 — Radian Measure, Coordinates at Key Angles, Reference Angles

The unit circle is a circle with radius 11 centered at the origin. Any point on the unit circle has coordinates:

(cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta)

This means:

  • cos⁡θ=\cos \theta = the xx-coordinate
  • sin⁡θ=\sin \theta = the yy-coordinate

Key angles and their coordinates:

θ\theta (rad)θ\theta (deg)(cos⁡θ,  sin⁡θ)(\cos \theta,\; \sin \theta)
000°0°(1,0)(1, 0)
π6\frac{\pi}{6}30°30°(32,12)\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)
π4\frac{\pi}{4}45°45°(22,22)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)
π3\frac{\pi}{3}60°60°(12,32)\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)
π2\frac{\pi}{2}90°90°(0,1)(0, 1)

Reference Angles

A reference angle is the acute angle formed between the terminal side and the xx-axis.

QuadrantReference angle formulasin⁡\sincos⁡\cos
I (0°0°–90°90°)θref=θ\theta_{\text{ref}} = \theta++++
II (90°90°–180°180°)θref=180°−θ\theta_{\text{ref}} = 180° - \theta++−-
III (180°180°–270°270°)θref=θ−180°\theta_{\text{ref}} = \theta - 180°−-−-
IV (270°270°–360°360°)θref=360°−θ\theta_{\text{ref}} = 360° - \theta−-++

Worked Example: Find cos⁡150°\cos 150°.

  1. 150°150° is in Quadrant II; reference angle =180°−150°=30°= 180° - 150° = 30°.
  2. cos⁡30°=32\cos 30° = \frac{\sqrt{3}}{2}.
  3. Cosine is negative in QII: cos⁡150°=−32\cos 150° = -\frac{\sqrt{3}}{2}.

Sign Patterns — "All Students Take Calculus"

A handy mnemonic for which trig functions are positive:

QuadrantPositive functionsMnemonic
IAll (sin⁡\sin, cos⁡\cos, tan⁡\tan)All
IIsin⁡\sin onlyStudents
IIItan⁡\tan onlyTake
IVcos⁡\cos onlyCalculus

Example: sin⁡210°\sin 210°

  1. QIII; reference angle =210°−180°=30°= 210° - 180° = 30°.
  2. sin⁡30°=12\sin 30° = \frac{1}{2}.
  3. Sine is negative in QIII: sin⁡210°=−12\sin 210° = -\frac{1}{2}.

Practice — Unit Circle 🔍

Find each value. Enter as a simplified fraction or integer. 🧮

  1. cos⁡0°=?\cos 0° = ?

  2. sin⁡270°=?\sin 270° = ?

  3. What is the reference angle for 315°315° (in degrees)?

Match each angle to its sine value. 🔍

SAT-Style Questions 📋

Part 5: Trig Applications

🎯 Trig on the SAT

Part 5 of 7 — Complementary Angles, Pythagorean Identity & SAT Favorites

Two high-frequency SAT trig concepts:

1. Complementary Angle Relationship:

If A+B=90°A + B = 90°, then:

sin⁡A=cos⁡Bandcos⁡A=sin⁡B\sin A = \cos B \qquad \text{and} \qquad \cos A = \sin B

This comes from the fact that in a right triangle, the two acute angles add to 90°90°, and one angle's opposite side is the other angle's adjacent side.

2. Pythagorean Identity:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This is always true — for every angle. Two useful rearrangements:

sin⁡2θ=1−cos⁡2θcos⁡2θ=1−sin⁡2θ\sin^2 \theta = 1 - \cos^2 \theta \qquad \cos^2 \theta = 1 - \sin^2 \theta

Complementary Angles — Worked Examples

Example 1: If sin⁡32°=0.53\sin 32° = 0.53, what is cos⁡58°\cos 58°?

Since 32°+58°=90°32° + 58° = 90°, they are complementary.

cos⁡58°=sin⁡32°=0.53\cos 58° = \sin 32° = 0.53

Example 2: In right triangle PQRPQR with a right angle at QQ, sin⁡P=45\sin P = \frac{4}{5}. What is cos⁡R\cos R?

Angles PP and RR are complementary (P+R=90°P + R = 90°), so: cos⁡R=sin⁡P=45\cos R = \sin P = \frac{4}{5}.


SAT Tip: If a question says "sin⁡x°=cos⁡y°\sin x° = \cos y°," immediately conclude x+y=90x + y = 90 (assuming 0<x,y<900 < x, y < 90).

Pythagorean Identity — Worked Examples

Example 1: If cos⁡θ=35\cos \theta = \frac{3}{5} and θ\theta is in Quadrant I, find sin⁡θ\sin \theta.

sin⁡2θ=1−cos⁡2θ=1−925=1625\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} sin⁡θ=45(positive in QI)\sin \theta = \frac{4}{5} \quad (\text{positive in QI})

Example 2: Simplify sin⁡2x+cos⁡2xcos⁡x\frac{\sin^2 x + \cos^2 x}{\cos x}.

=1cos⁡x= \frac{1}{\cos x}

The numerator is exactly 11 by the identity, so the whole expression collapses.

Example 3: If sin⁡θ=0.6\sin \theta = 0.6, what is sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta?

=1(always, by identity)= 1 \quad \text{(always, by identity)}

The SAT might try to distract you — don't compute; just recognize the identity.

Practice — SAT Trig Concepts 🔍

Compute each value. 🧮

  1. If sin⁡(2x)°=cos⁡(x+15)°\sin(2x)° = \cos(x + 15)° and both angles are acute, what is xx?

  2. If cos⁡θ=0.8\cos \theta = 0.8, what is sin⁡2θ\sin^2 \theta? (Enter as a decimal.)

  3. sin⁡240°+cos⁡240°=?\sin^2 40° + \cos^2 40° = ?

Match each expression to its equivalent. 🔍

SAT-Style Questions 📋

Part 6: Problem-Solving Workshop

🔶 Circle Theorems & Tangent Lines

Part 6 of 7 — Inscribed Angles, Central Angles, Tangent-Radius Perpendicularity

Key circle theorems tested on the SAT:

TheoremStatement
Central angleCentral angle == intercepted arc
Inscribed angleInscribed angle =12= \frac{1}{2} intercepted arc
Tangent-radiusA tangent line is ⊥\perp to the radius at the point of tangency
Two tangentsTangent segments from an external point are equal in length

Central Angle vs. Inscribed Angle:

If a central angle and an inscribed angle intercept the same arc, the central angle is twice the inscribed angle.

Central angle=2×Inscribed angle\text{Central angle} = 2 \times \text{Inscribed angle}

Inscribed Angles — Worked Examples

Example 1: A central angle measures 80°80°. An inscribed angle intercepts the same arc. What is the inscribed angle?

Inscribed=80°2=40°\text{Inscribed} = \frac{80°}{2} = 40°

Example 2: An inscribed angle in a semicircle?

Any angle inscribed in a semicircle intercepts a 180°180° arc.

Inscribed angle=180°2=90°\text{Inscribed angle} = \frac{180°}{2} = 90°

This is Thales' theorem: an angle inscribed in a semicircle is always a right angle.

Example 3: Two inscribed angles intercept the same arc. What can you say?

They are equal — inscribed angles that intercept the same arc are congruent.


SAT Tip: If you see a triangle inscribed in a circle with one side as a diameter, immediately mark a 90°90° angle.

Tangent Lines — Key Properties

A tangent touches the circle at exactly one point and is perpendicular to the radius at that point.

Worked Example: Point PP is outside a circle with center OO and radius 55. A tangent from PP touches the circle at TT, and OP=13OP = 13. Find PTPT.

  1. OT⊥PTOT \perp PT (tangent-radius), so triangle OTPOTP is a right triangle.
  2. OT=5OT = 5 (radius), OP=13OP = 13 (given).
  3. By the Pythagorean theorem: PT=132−52=169−25=144=12PT = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.

Two tangents from one point: If two tangent segments are drawn from the same external point, they have equal length.

So if PAPA and PBPB are tangent to a circle at AA and BB, then PA=PBPA = PB.

Practice — Circle Theorems 🔍

Solve each problem. 🧮

  1. A central angle measures 110°110°. What is the minor arc it intercepts (in degrees)?

  2. An inscribed angle measures 55°55°. What is the intercepted arc (in degrees)?

  3. Triangle ABCABC is inscribed in a circle with BCBC as a diameter. What is angle AA (in degrees)?

Match each scenario to the correct conclusion. 🔍

SAT-Style Questions 📋

Part 7: Review & Applications

🏁 Review & Mixed Practice

Part 7 of 7 — Formula Cheat Sheet, Mixed Problems & Exam Strategies

Formula Cheat Sheet

TopicFormula
Circle equation(x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2
Arc lengths=rθs = r\theta
Sector areaA=12r2θA = \frac{1}{2}r^2\theta
sin⁡\sinoppositehypotenuse\frac{\text{opposite}}{\text{hypotenuse}}
cos⁡\cosadjacenthypotenuse\frac{\text{adjacent}}{\text{hypotenuse}}
tan⁡\tanoppositeadjacent\frac{\text{opposite}}{\text{adjacent}}
Pythagorean identitysin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
Complementary anglessin⁡A=cos⁡(90°−A)\sin A = \cos(90° - A)
Inscribed angle=12= \frac{1}{2} intercepted arc
Tangent-radiusPerpendicular (90°90°)

Special triangles:

45-45-9045\text{-}45\text{-}901:1:21 : 1 : \sqrt{2}
30-60-9030\text{-}60\text{-}901:3:21 : \sqrt{3} : 2

SAT Exam Strategies for Circles & Trig

1. Read the question twice. Common traps:

  • Asking for r2r^2 vs. rr
  • Giving degrees when you need radians (or vice-versa)
  • Asking for an angle when you solve for a side

2. Draw and label. Sketch the circle or triangle directly on the test — label known sides, angles, and what you're solving for.

3. Look for right triangles. Tangent-radius? That's a right angle. Diameter as a chord? Thales gives you 90°90°.

4. Use process of elimination. Trig ratios are between −1-1 and 11 (for sin⁡\sin and cos⁡\cos). If an answer is >1> 1, it can't be a sine or cosine value.

5. Pythagorean triples save time: 3-4-53\text{-}4\text{-}5, 5-12-135\text{-}12\text{-}13, 8-15-178\text{-}15\text{-}17, 7-24-257\text{-}24\text{-}25 and their multiples.

Mixed Practice Set 1 🔍

Mixed Practice Set 2 🔍

Final mixed problems — enter your answers. 🧮

  1. What is the area of a circle with equation (x+1)2+(y−3)2=49(x+1)^2 + (y-3)^2 = 49? Enter in terms of π\pi — give just the coefficient (e.g., if the answer is 25π25\pi, enter 25).

  2. A tangent to a circle of radius 77 is drawn from a point 2525 units from the center. How long is the tangent segment?

  3. cos⁡272°+sin⁡272°=?\cos^2 72° + \sin^2 72° = ?

Quick-fire review — choose the correct value. 🔍

📋 What You've Mastered

Across all 7 parts you've covered:

  • ✅ Circle equations — standard form, completing the square, center & radius
  • ✅ Arc length & sector area with degree and radian formulas
  • ✅ Right triangle trig — SOH-CAH-TOA and special triangles
  • ✅ The unit circle — coordinates, reference angles, sign patterns
  • ✅ SAT trig favorites — complementary angles and Pythagorean identity
  • ✅ Circle theorems — inscribed angles, central angles, tangent-radius
  • ✅ Mixed practice and exam strategies

Next steps:

  • Time yourself: aim for each circle/trig problem in under 90 seconds.
  • On test day, sketch diagrams — even rough drawings reveal right triangles and relationships.
  • If stuck, try plugging in the answer choices. For trig questions, use special values (30°30°, 45°45°, 60°60°) to test.

Good luck on the SAT! 🚀