ORIGINAL PRACTICE · 1
A circle has circumference 18π cm. Find its radius and area.
Reveal answer & reasoning
Radius 9 cm; area 81π cm²
From 2πr = 18π, r = 9. Then πr² = 81π. The circumference determines a length first; it is not itself an area.
SAT · Geometry and Trigonometry · TOPIC LESSON
Circles
Connect circle geometry with radians, coordinate equations, and the unit circle.
BY THE END, YOU CAN
- Use radius, diameter, chord, tangent, and intercepted-arc relationships.
- Calculate arc length, sector area, and degree–radian conversions.
- Create and interpret circle equations, including by completing the square.
- Use unit-circle coordinates, quadrant signs, and coterminal angles.
BEFORE YOU START
- Use perimeter and area formulas and simplify expressions with π.
- Apply the Pythagorean theorem, coordinate distance, and midpoint formulas.
- Expand squared binomials, solve simple quadratic equations, and recognize special right triangles.
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original SAT-style worked examples and practice
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
A circle is a constant-distance set
A circle is the set of points in a plane at a fixed positive distance r from a center. That distance is the radius. A diameter passes through the center and has length 2r. A chord joins any two points on the circle; a diameter is the longest chord.
A secant line intersects the circle in two points; a tangent line meets it in one point. A circular arc is part of the boundary, while a sector is the region between two radii and an arc. Circumference measures the complete boundary; disk area measures the enclosed region.
Distinguish arc measure from arc length. An arc can have angular measure 80° but length 5 cm. Angles are compared with a full turn, whereas lengths depend on the circle’s radius. Two circles can have equal angular arcs but different arc lengths.
d = 2r; C = 2πr = πd; A = πr²
Radius and diameter have length units; circumference has length units; area has square units.
02 / UNDERSTAND & APPLY
Locate the angle’s vertex and intercepted arc
A central angle has its vertex at the center. Its degree measure equals the degree measure of the arc it intercepts. An inscribed angle has its vertex on the circle and sides along chords; its measure is half the measure of the intercepted arc that does not contain its vertex.
The intercepted arc is essential. Two inscribed angles subtending the same arc are equal. If their vertices lie on opposite sides of the same chord, they intercept different arcs, and their measures are supplementary rather than equal.
An angle inscribed in a semicircle is 90° because it intercepts a 180° arc. Opposite angles in a cyclic quadrilateral sum to 180° because their intercepted arcs together form a full 360° circle. Use these consequences only when the stated points actually lie on the same circle.
Inscribed angle = 1/2 × intercepted arc measure
The intercepted arc is the one between the chord endpoints that excludes the angle’s vertex.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
One chord, two possible arcs
The minor arc AB measures 124°. Point P lies on the major arc AB, and point Q lies on the minor arc AB. Find ∠APB and ∠AQB.
- ∠APB intercepts the minor arc AB, because that arc does not contain P. Its measure is 124/2 = 62°.
- The major arc AB measures 360 − 124 = 236°.
- ∠AQB intercepts that major arc, so its measure is 236/2 = 118°. The two angles sum to 180°.
Answer ∠APB = 62°; ∠AQB = 118°
ORIGINAL PRACTICE · 1
AB is a diameter of a circle, and C is another point on the circle. If ∠CAB = 28°, find ∠ABC.
Reveal answer & reasoning
62°
The angle at C intercepts diameter AB, so it is 90°. The triangle’s other two angles sum to 90°, giving ∠ABC = 90 − 28 = 62°.
ORIGINAL PRACTICE · 2
An inscribed quadrilateral has one angle of 107°. Find its opposite angle.
Reveal answer & reasoning
73°
Opposite angles of an inscribed quadrilateral are supplementary. Subtract 107 from 180 to get 73. Adjacent angles are not generally supplementary.
03 / UNDERSTAND & APPLY
Use a perpendicular from the center to a chord
A perpendicular from the center of a circle to a chord bisects the chord. Joining the center to the two endpoints creates two right triangles with equal hypotenuses (radii) and a shared leg, so the remaining legs are equal.
If the center-to-chord distance is d and the radius is r, half the chord has length √(r² − d²). The full chord is twice that. A chord closer to the center is longer; d = 0 gives a diameter, the longest possible chord.
Equal chords in the same circle are the same distance from the center and subtend equal minor central angles. The same-circle condition matters: equal chords in circles of different radii need not have equal center distances.
Chord length = 2√(r² − d²)
d is the perpendicular distance from the center to the chord, with 0 ≤ d < r for a nondegenerate chord.
ORIGINAL WORKED EXAMPLE
A chord from its distance to the center
A circle has radius 13 cm. A chord is 5 cm from the center, measured perpendicularly. Find the chord’s length.
- The perpendicular from the center meets the chord at its midpoint.
- Half the chord forms a right triangle with leg 5 and hypotenuse 13.
- Half-chord length = √(13² − 5²) = √144 = 12 cm.
- Double the half-chord: the full chord is 24 cm.
Answer 24 cm
ORIGINAL PRACTICE · 1
A chord of a circle with radius 10 has length 16. Find its perpendicular distance from the center.
Reveal answer & reasoning
6
The half-chord is 8. The radius, half-chord, and center distance form a right triangle: d² + 8² = 10², so d² = 36 and d = 6. Using the full 16 as a leg would be impossible.
04 / UNDERSTAND & APPLY
A radius is perpendicular to a tangent at contact
At the point of tangency T, radius OT is perpendicular to the tangent line. This right angle can turn a circle problem into a right-triangle problem. The perpendicular relation applies at the point of contact; a radius drawn to some other point need not be perpendicular to that tangent.
If two tangents are drawn from the same exterior point, their tangent segments from that point to the two contact points have equal lengths. The two right triangles share the center-to-exterior-point hypotenuse and have equal radii, establishing congruence.
The angle between a tangent and a chord equals half the measure of the arc intercepted by that angle. There are supplementary choices of tangent–chord angle, corresponding to the minor and major arcs. Identify the actual angle and arc instead of automatically halving the minor arc.
ORIGINAL WORKED EXAMPLE
Find the length to the contact point
O is a circle’s center, P is outside the circle, and PT is tangent at T. If OP = 17 and OT = 8, find PT.
- OT is a radius to the contact point, so OT is perpendicular to PT.
- OP is the hypotenuse of right triangle OTP.
- PT² = OP² − OT² = 17² − 8² = 225.
- The positive tangent length is PT = 15.
Answer 15
ORIGINAL PRACTICE · 1
Two tangent segments from exterior point P have lengths 3x + 1 and 5x − 9. Find each length.
Reveal answer & reasoning
16
The two tangent lengths from one exterior point are equal: 3x + 1 = 5x − 9. Thus x = 5, and each expression gives 16.
ORIGINAL PRACTICE · 2
A tangent and chord meet at T. The angle between the specified tangent ray and chord intercepts an arc measuring 146°. Find that angle.
Reveal answer & reasoning
73°
A tangent–chord angle has half the measure of its intercepted arc: 146/2 = 73°. The supplementary tangent ray would form 107°, so specifying the ray and intercepted arc matters.
05 / UNDERSTAND & APPLY
Arc length is a fraction of circumference
A central angle of θ degrees selects the fraction θ/360 of a full turn. The corresponding arc length is the same fraction of the circumference. The formula applies to major arcs too if θ is the matching angle between 180° and 360°.
If an inscribed angle is given, double it to find the intercepted arc measure before using the circumference fraction. A chord length is a straight-line distance, so it is not equal to its arc length except in the limiting sense for an extremely small arc.
Solve backward when arc length is given. Dividing the arc length by the whole circumference gives the fraction of a turn; multiply that fraction by 360 to obtain degrees. Use the radius of the same circle throughout.
s = (θ/360°) · 2πr
s is arc length and θ is its central angle in degrees.
ORIGINAL WORKED EXAMPLE
An exact arc length
A circle has radius 12 cm. Find the length of the arc intercepted by a 75° central angle.
- The arc is 75/360 = 5/24 of the circle.
- The full circumference is 2π(12) = 24π cm.
- Multiply: s = (5/24)(24π) = 5π cm.
Answer 5π cm
ORIGINAL PRACTICE · 1
A circle has radius 10. An inscribed angle measures 54°. Find the length of its intercepted arc.
Reveal answer & reasoning
6π
The intercepted arc measures 2 × 54 = 108°. Its length is (108/360)(20π) = 6π. Using 54° directly would give half the correct length.
06 / UNDERSTAND & APPLY
Radians measure arc length relative to radius
One radian is the central angle that intercepts an arc as long as the radius. In general, θ = s/r, so an angle in radians is a ratio of two lengths. A full circle has arc length 2πr and therefore 2π radians.
Since 180° = π radians, multiply a degree measure by π/180 to convert to radians, and a radian measure by 180/π to convert to degrees. An angle of π radians is 180°, not 3.14°. Keep the unit attached when moving between systems.
For an angle in radians, arc length simplifies to s = rθ. This formula does not accept a degree number directly. An angle of 60° must first become π/3 radians before multiplying by the radius.
θradians = θdegrees · π/180; θdegrees = θradians · 180/π
Multiply by the conversion factor that cancels the old angle unit.
s = rθ
θ must be in radians; radius and arc length use the same length unit.
ORIGINAL WORKED EXAMPLE
Convert and use a radian angle
A circle has radius 6 m. A central angle measures 5π/6 radians. Find its degree measure and arc length.
- Degrees = (5π/6)(180/π) = 150°.
- The angle is already in radians, so s = rθ = 6(5π/6) = 5π m.
- As a check, 150° is 5/12 of a full turn, and (5/12)(12π) = 5π.
Answer 150°; arc length 5π m
ORIGINAL PRACTICE · 1
Convert 225° to radians.
Reveal answer & reasoning
5π/4 radians
Multiply by π/180: 225π/180 simplifies to 5π/4. Because 225° is between 180° and 270°, the result should lie between π and 3π/2.
ORIGINAL PRACTICE · 2
An arc has length 14 cm in a circle of radius 8 cm. Find the central angle in radians.
Reveal answer & reasoning
7/4 radians
Use θ = s/r = 14/8 = 7/4. There is no extra π unless it appears in the given length or the equivalent exact angle. The ratio of lengths is dimensionless.
07 / UNDERSTAND & APPLY
Sector area and sector perimeter answer different questions
A sector’s area is the same fraction of the disk area as its central angle is of a full turn. Use θ/360 for degrees or θ/(2π) for radians. With radians, the expression simplifies to A = r²θ/2.
The perimeter of an ordinary sector with an angle strictly between 0 and a full turn consists of the selected arc plus two radii. Sector area does not include the triangle’s area as an extra term; the sector already contains it. A circular segment, in contrast, is the region between an arc and its chord, often found as sector area minus triangle area.
Arc length and sector area can be linked directly: because s = rθ, A = rs/2. This is useful when an arc length and radius are given but no angle is stated. Check units: r times s has square units.
Asector = (θ/360°)πr² = r²θradians/2
Use the version that matches the angle unit.
Asector = rs/2; Psector = s + 2r
The perimeter formula is for a non-full, nonzero sector bounded by two distinct radii.
ORIGINAL WORKED EXAMPLE
Area and perimeter of a sector
A sector of a radius-9 cm circle has central angle 80°. Find its area and perimeter.
- The fraction of a full circle is 80/360 = 2/9.
- Area = (2/9)π(9²) = 18π cm².
- Arc length = (2/9)(2π × 9) = 4π cm.
- Perimeter includes both radii: 4π + 9 + 9 = 18 + 4π cm.
Answer Area 18π cm²; perimeter 18 + 4π cm
ORIGINAL PRACTICE · 1
A sector has radius 5 and arc length 8. Find its area.
Reveal answer & reasoning
20 square units
Use A = rs/2 = (5)(8)/2 = 20. Alternatively θ = 8/5 radians and A = (1/2)(25)(8/5) = 20.
ORIGINAL PRACTICE · 2
A 90° sector of a radius-4 circle contains the triangle formed by its two radii and chord. Find the area between that chord and the arc.
Reveal answer & reasoning
4π − 8 square units
The sector is one quarter of the disk, so its area is (1/4)π(4²) = 4π. The included triangle is right with legs 4 and 4, so its area is 8. Subtract to get the circular segment area 4π − 8.
08 / UNDERSTAND & APPLY
Read and create the equation of a circle
A point (x, y) lies on a circle centered at (h, k) with radius r exactly when its distance from the center is r. Squaring the distance formula gives (x − h)² + (y − k)² = r². The right side is radius squared, not radius.
Read the center by the quantities subtracted inside the parentheses: (x + 2)² means h = −2. Horizontal and vertical extreme points are (h ± r, k) and (h, k ± r). These help sketch a circle with equal scale on both axes.
Translating a circle adds the horizontal and vertical shifts to its center coordinates and leaves its radius unchanged. Resizing about its center keeps the center fixed but multiplies the radius by the positive scale factor. The right side of the equation therefore multiplies by the square of that factor, not by the factor itself.
If a diameter’s endpoints are given, the center is their midpoint and the radius is half their distance apart. If the center and one point are given, calculate r² directly from the coordinate differences; there is no need to find a square root only to square it again.
(x − h)² + (y − k)² = r²
Center (h, k); radius r > 0.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Build a circle from a diameter
A circle’s diameter has endpoints (−4, 1) and (8, 5). Find its equation.
- The midpoint is ((−4 + 8)/2, (1 + 5)/2) = (2, 3).
- From center (2, 3) to endpoint (8, 5), the squared distance is 6² + 2² = 40.
- Thus r² = 40 and the equation is (x − 2)² + (y − 3)² = 40.
Answer (x − 2)² + (y − 3)² = 40
ORIGINAL WORKED EXAMPLE
Translate and enlarge a circle
The circle (x + 2)² + (y − 3)² = 16 is translated 5 units right and 2 units down. Its radius is then multiplied by 3/2 about the new center. Find the new equation.
- The original center is (−2, 3), and radius is 4.
- Translation changes the center to (−2 + 5, 3 − 2) = (3, 1) without changing the radius.
- The new radius is 4(3/2) = 6, so the new squared radius is 36.
- The new equation is (x − 3)² + (y − 1)² = 36.
Answer (x − 3)² + (y − 1)² = 36
ORIGINAL PRACTICE · 1
A circle with equation (x − 4)² + (y + 1)² = 25 is shifted 3 units left and has its radius doubled about the shifted center. Find the new equation.
Reveal answer & reasoning
(x − 1)² + (y + 1)² = 100
The center moves from (4, −1) to (1, −1). The radius grows from 5 to 10, so the right side grows from 25 to 100, a factor of 4. The vertical center coordinate does not change.
ORIGINAL PRACTICE · 2
A circle has equation (x + 2)² + (y − 3)² = 16. Find its center, radius, and rightmost point.
Reveal answer & reasoning
Center (−2, 3); radius 4; rightmost point (2, 3)
The center is (−2, 3) because the x term is x − (−2). The radius is √16 = 4. Add 4 to the center’s x-coordinate for the rightmost point.
ORIGINAL PRACTICE · 3
A circle centered at (4, −1) passes through (−2, 7). Find its equation.
Reveal answer & reasoning
(x − 4)² + (y + 1)² = 100
The coordinate differences from center to point are −6 and 8, so r² = 36 + 64 = 100. Substitute center (4, −1) with the correct signs into standard form.
09 / UNDERSTAND & APPLY
Recover center and radius from expanded form
An expanded circle equation hides the center in the linear terms. Group the x terms and y terms, move the constant, and complete each square. To turn x² + bx into a perfect square, add (b/2)². Whatever is added on one side must also be added on the other.
If x² and y² have the same nonzero coefficient, divide the entire equation by that coefficient first. Unequal squared-term coefficients generally describe a different conic, not a circle in the usual Cartesian coordinates. Do not complete squares as though the leading coefficient were 1 without normalizing.
After completing squares, interpret the right side. A positive value gives a circle with its positive square-root radius. A zero value gives only the center point; a negative value gives no real points. It is not meaningful to report a negative radius.
ORIGINAL WORKED EXAMPLE
Normalize, complete, interpret
Find the center and radius of 2x² + 2y² − 12x + 8y − 24 = 0.
- Divide every term by 2: x² + y² − 6x + 4y − 12 = 0.
- Group and move the constant: (x² − 6x) + (y² + 4y) = 12.
- Add 9 and 4 to both sides: (x − 3)² + (y + 2)² = 25.
- The center is (3, −2), and the radius is √25 = 5.
Answer Center (3, −2); radius 5
ORIGINAL PRACTICE · 1
Find the center and radius of x² + y² + 8x − 10y + 32 = 0.
Reveal answer & reasoning
Center (−4, 5); radius 3
Move 32 to the right and add 16 and 25: (x + 4)² + (y − 5)² = −32 + 16 + 25 = 9. Thus the radius is 3, not 9.
ORIGINAL PRACTICE · 2
Does x² + y² − 2x + 6y + 14 = 0 describe a real circle?
Reveal answer & reasoning
No real points
Completing the squares gives (x − 1)² + (y + 3)² = −4. A sum of real squares cannot be negative, so there is no real circle or point.
10 / UNDERSTAND & APPLY
Use equations to test points and find intersections
For a circle centered at (h, k), compare a point’s squared distance (x − h)² + (y − k)² with r². Equality means on the circle, a smaller value means inside the disk, and a larger value means outside. The circle itself is the boundary; an inequality can describe the disk.
To find where a line intersects a circle, substitute the line’s relationship into the circle equation. Two real solutions usually give a secant, one repeated real solution gives tangency, and no real solutions mean the line misses the circle. A horizontal or vertical line makes this especially direct.
A graph is a useful check, but an uneven axis scale can make a circle appear elliptical. Exact coordinate algebra establishes tangency more reliably than a screen image that merely looks tangent. At a known tangent point, the tangent direction is perpendicular to the radius direction.
ORIGINAL WORKED EXAMPLE
A secant and a tangent from the same equation
For the circle x² + y² = 25, find its intersections with y = 3 and with x = 5.
- For y = 3, substitute to get x² + 9 = 25, so x² = 16 and x = ±4. The intersections are (−4, 3) and (4, 3).
- For x = 5, substitute to get 25 + y² = 25, so y = 0. There is one contact point, (5, 0).
- The radius to (5, 0) is horizontal; the tangent line x = 5 is vertical, confirming perpendicularity.
Answer y = 3 intersects at (−4, 3) and (4, 3); x = 5 is tangent at (5, 0).
ORIGINAL PRACTICE · 1
Is (1, 5) inside, on, or outside the circle (x + 2)² + (y − 3)² = 16?
Reveal answer & reasoning
Inside the disk
Its squared distance to (−2, 3) is (1 + 2)² + (5 − 3)² = 9 + 4 = 13, which is less than 16. It is inside the enclosed disk, not on the circle boundary.
ORIGINAL PRACTICE · 2
Find the horizontal tangent lines to (x − 1)² + (y + 4)² = 9.
Reveal answer & reasoning
y = −1 and y = −7
The center is (1, −4) and radius is 3. Horizontal tangents occur at the top and bottom, with y = −4 + 3 = −1 and y = −4 − 3 = −7.
11 / UNDERSTAND & APPLY
Extend trig ratios with unit-circle coordinates
The unit circle has center (0, 0), radius 1, and equation x² + y² = 1. An angle in standard position starts at the positive x-axis; positive angles turn counterclockwise. The endpoint on the unit circle is (cos θ, sin θ). This agrees with right-triangle ratios in the first quadrant because the radius is the hypotenuse of length 1.
Coordinate signs determine trig signs. In quadrant I both cosine and sine are positive; in quadrant II cosine is negative and sine positive; in quadrant III both are negative; in quadrant IV cosine is positive and sine negative. Tangent is y/x, so it is undefined where x = 0.
Use a reference angle, the acute angle between the terminal ray and the nearest x-axis direction, to get magnitudes from a special triangle, then apply signs. At π/6, the point is (√3/2, 1/2); at π/4, it is (√2/2, √2/2); at π/3, it is (1/2, √3/2).
Angles differing by an integer multiple of 2π radians or 360° are coterminal and reach the same point. Reducing an angle by complete turns simplifies it without changing sine or cosine. Negative angles turn clockwise. The identity sin²θ + cos²θ = 1 follows from the circle equation for every real θ, with signs still determined separately.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A second-quadrant angle
Find the unit-circle point at θ = 5π/6 and the values of sin θ, cos θ, and tan θ.
- 5π/6 = 150°, so the terminal point is in quadrant II. Its reference angle is π/6.
- For π/6 the coordinate magnitudes are √3/2 and 1/2. Quadrant II makes x negative and y positive.
- The point is (−√3/2, 1/2), so cosine is −√3/2 and sine is 1/2.
- Tangent is y/x = (1/2)/(−√3/2) = −1/√3 = −√3/3.
Answer Point (−√3/2, 1/2); sin θ = 1/2; cos θ = −√3/2; tan θ = −√3/3
ORIGINAL PRACTICE · 1
Find the unit-circle point for angle 13π/6.
Reveal answer & reasoning
(√3/2, 1/2)
Subtract a full turn: 13π/6 − 2π = π/6. Coterminal angles have the same endpoint, and the π/6 point is (√3/2, 1/2).
ORIGINAL PRACTICE · 2
A point on the unit circle lies in quadrant IV and has x-coordinate 3/5. Find its y-coordinate.
Reveal answer & reasoning
−4/5
The equation x² + y² = 1 gives y² = 1 − 9/25 = 16/25. The square-root magnitudes are 4/5, but quadrant IV requires a negative y-coordinate.
ORIGINAL PRACTICE · 3
At θ = 3π/2, find sine, cosine, and tangent.
Reveal answer & reasoning
sin θ = −1; cos θ = 0; tan θ is undefined
The unit-circle point is (0, −1). Sine is the y-coordinate and cosine is the x-coordinate. Tangent would be −1/0, which is undefined.
ORIGINAL PRACTICE · 4
Find cos(−π/3) and sin(−π/3).
Reveal answer & reasoning
cos(−π/3) = 1/2; sin(−π/3) = −√3/2
A negative angle turns clockwise from the positive x-axis. The endpoint lies in quadrant IV with reference angle π/3, so x is positive 1/2 and y is negative √3/2.
KEEP THE BIG IDEAS
Quick revision
- Determine whether an angle is central, inscribed, or tangent–chord before using its arc.
- A radius is perpendicular to a tangent at the contact point.
- Arc length and sector area use the same fraction of a full turn.
- Use s = rθ and A = r²θ/2 only with θ in radians.
- In a circle equation, the center has the opposite signs from the constants inside the parentheses and the right side is r².
- Normalize equal squared coefficients before completing the square, and check that r² is positive.
- On the unit circle, coordinates are (cos θ, sin θ); reference angles give magnitudes and quadrants give signs.
READ WITH CONTEXT
References & scope
The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.
- College Board: SAT Geometry and Trigonometry ↗
Official domain outline; used to check topic placement. All teaching, worked examples, and practice in this lesson are original.
- College Board: Math skill descriptions ↗
The public skill descriptions establish the topic scope. No question-bank questions or explanations are reproduced.
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