Math 2 — Quiz Review 12.0–12.2A — Fully Revised Answer Key

1

Circle Parts — center P; points J, K, L, M, N on circle

Give an example of each circle part from the diagram.

📐 FoundationA circle is the set of ALL points equidistant from one fixed center point. Every circle term is defined by its relationship to the center.

a. Center → P

The single interior point from which every point on the circle is the same distance (= radius).

b. Radius → LP, MP, or NP

Any segment from center P to a point on the circle. All radii of a circle are congruent — that is part of the definition.

c. Diameter → any chord through P

Longest chord, passes through the center. Diameter = 2 × radius. If you know one, you can always find the other.

d. Chord → JK, KL, or JM

Any segment with BOTH endpoints on the circle. Does not need to pass through the center. A diameter is just the special chord that does.

e. Secant → the full LINE through any chord

A secant is an infinite line that intersects the circle at exactly 2 points. The chord is just the segment between those 2 points.

f. Tangent → a line touching the circle at exactly 1 point

Always perpendicular to the radius drawn to the point of tangency. This 90° relationship is the key to all tangent problems.

g. Central Angle → ∠LPM (vertex AT center P)

Central Angle Theorem: central angle = intercepted arc. They have the exact same degree measure.

h. Inscribed Angle → ∠JKL (vertex ON the circle)

Inscribed Angle Theorem: inscribed angle = ½ × intercepted arc. Always half the central angle intercepting the same arc.

i. Minor Arc → arc JK (< 180°, named with 2 letters)

The shorter arc between two points.

j. Major Arc → arc JMK (> 180°, named with 3 letters)

The longer arc. The middle letter is any point on that arc, used to distinguish from the minor arc.

k. Semicircle → exactly 180°

Endpoints are the ends of a diameter. Thales' Theorem: an inscribed angle intercepting a semicircle always = 90°.

2

Arc Measures — center H; diameter GE; chord CD ⊥ GE at H; arc FE = 34°

Find mCD, mFD, mDCF, mGDF.

📐 TheoremCentral Angle Theorem: arc = central angle. All arcs in a circle sum to 360°.

⚠ Common MistakeArc ED = 90° on its own from the right-angle mark at H. Do NOT compute 90° − 34° = 56° for arc ED. Arc FE = 34° and arc ED = 90° are two separate, adjacent arcs. You ADD them, you do not subtract.

Setup: GE is a diameter through center H → the circle is split into two semicircles of 180° each.
Right-angle square at H shows ∠EHD = 90° → Central Angle Theorem → arc ED = 90°.
Bottom semicircle: arc GF + arc FE + arc ED = 180° → arc GF + 34° + 90° = 180° → arc GF = 56°.
Top: vertical angle ∠GHC = 90° → arc GC = 90°, arc CD = 90°.

mCD

Arc CD is intercepted by central angle ∠CHD = 90°.

mCD = 90°

mFD

Going from F to D the short way: F → E → D.
arc FE = 34° (given), arc ED = 90° (from the right-angle mark — its own arc).
mFD = 34° + 90° = 124°.

⚠ CorrectionPrevious versions of this document incorrectly stated mFD = 90°. The correct answer is 124°.

mFD = 124°

mDCF — major arc D → C → G → F (the long way, not through E)

arc DC = 90°, arc CG = 90°, arc GF = 56°.
mDCF = 90° + 90° + 56° = 236°.

💡 Check: minor arc FD (124°) + major arc DCF (236°) = 360° ✓

⚠ CorrectionPrevious versions stated mDCF = 270°. The correct answer is 236°.

mDCF = 236°

mGDF — arc G → F → E → D (going clockwise along the bottom)

arc GF = 56°, arc FE = 34°, arc ED = 90°.
mGDF = 56° + 34° + 90° = 180°.

💡 G and D are endpoints of a diameter (GE is a diameter, and D sits directly perpendicular to E through center H). Arc GD through the bottom = 180° confirms this.

mGDF = 180°

3

Arc Measures — center U; 19° central angle; PR and QS are diameters

Find mPQ, mSR, mQRT, mPSR, mPS.

📐 TheoremCentral angle = intercepted arc. Vertical angles at center → equal arcs across the circle. Diameter → 180° arc.

Two chords (diameters) cross at center U. Central angle ∠PUQ = 19° → arc PQ = 19°.

mPQ

mPQ = 19°

mSR

∠SUR is vertical to ∠PUQ → equal → arc SR = 19°.

mSR = 19°

mQRT

arc QR: PR is a diameter → arc PR = 180°. arc PQ = 19° → arc QR = 180° − 19° = 161°.
arc RT: vertical to arc PQ → arc RT = 19°.
mQRT = 161° + 19° = 180°.

💡 QT is also a diameter → arc = 180° ✓

mQRT = 180°

mPSR

Arc PS (not through Q) = 180° − 19° = 161°. Arc SR = 19°.
mPSR = 161° + 19° = 180°.

💡 PR is a diameter → arc P to R = 180° ✓

mPSR = 180°

mPS

Going from P to S (not through Q): 180° − arc PQ = 180° − 19° = 161°.

mPS = 161°

4

Arc Length — radius = 7.5 in, central angle = 35°

Find the arc length. Round to nearest tenth.

📐 FormulaArc Length = (angle / 360°) × 2πr

💡 An arc is just a fraction of the full circumference. That fraction = angle ÷ 360.

= (35 / 360) × 2π × 7.5
= (35 / 360) × 47.1239...
= 0.09722 × 47.1239 = 4.581...

Arc Length ≈ 4.6 inches

5

Find Radius — central angle = 60°, arc length = 5 m

Find the radius. (No diagram provided.)

📐 FormulaArc Length = (angle / 360) × 2πr → isolate r: r = (Arc Length × 360) / (2π × angle)

5 = (60 / 360) × 2πr = (1/6) × 2πr = πr/3
15 = πr
r = 15/π = 4.7746...

r ≈ 4.8 meters

6

Sector Area — radius JI = 15.1 in, central angle = 72°

Find the area of the sector. Round to nearest tenth.

📐 FormulaSector Area = (angle / 360°) × πr²

⚠ Read CarefullyJI = 15.1 in is the RADIUS (from center J to point I on the circle). Do not use it as diameter.

= (72 / 360) × π × (15.1)²
= (1/5) × π × 228.01
= (1/5) × 716.283 = 143.26

Sector Area ≈ 143.3 in²

7

Find Central Angle — r = 12 cm, sector area = 32 cm²

Find the central angle. (No diagram provided.)

📐 Formulaθ = (Area × 360) / (πr²)

32 = (θ / 360) × π × 144
θ/360 = 32 / 452.389 = 0.070736
θ = 0.070736 × 360 = 25.47°

💡 Check: (25.47/360) × π × 144 ≈ 32 cm² ✓

Central Angle ≈ 25.5°

8

Shaded Area — JM = 10 mm is the DIAMETER; unshaded sector = 58°

Find the total area of the shaded region. Round to nearest tenth.

📐 StrategyShaded Area = Total Circle Area − Unshaded Sector Area

⚠ Key DetailJM = 10 mm is the DIAMETER → radius = 10 ÷ 2 = 5 mm. Using 10 as the radius would give a wrong answer 4× too large.

radius = 5 mm
Total area = π × 5² = 25π = 78.540 mm²
Unshaded sector = (58 / 360) × 78.540 = 12.654 mm²
Shaded = 78.540 − 12.654 = 65.886 mm²

💡 Verify: shaded angle = 360°−58° = 302°. (302/360) × 78.540 = 65.89 ✓

Shaded Area ≈ 65.9 mm²

9

Shaded Area — PC = 12 cm is the RADIUS; three shaded sectors of 62° each

Find the total area of the shaded region. Round to nearest tenth.

📐 StrategyTotal Shaded Area = (combined shaded angle / 360°) × πr²

⚠ Key DetailPC = 12 cm: P is the center, C is on the circle → PC = radius = 12 cm. The 6 sectors alternate shaded/unshaded. Three shaded at 62° each; three unshaded at (360°−186°)/3 = 58° each.

Total shaded angle = 3 × 62° = 186°
Circle area = π × 12² = 144π = 452.389 cm²
Shaded = (186 / 360) × 452.389 = 0.5167 × 452.389 = 233.74 cm²

Shaded Area ≈ 233.7 cm²

10

Is AB Tangent to Circle P? — PA = 8, AB = 15, PB = 17

Determine if segment AB is tangent to circle P.

📐 TheoremIf AB is tangent at A, then PA ⊥ AB → right angle at A → must satisfy a² + b² = c².

PA = 8 (radius), AB = 15 (potential tangent), PB = 17 (hypotenuse — always the longest side).
8² + 15² = 64 + 225 = 289
17² = 289
289 = 289 ✓

💡 8-15-17 is a Pythagorean triple. Right angle confirmed at A → PA ⊥ AB.

YES — AB is tangent to circle P.

11

Is AB Tangent to Circle P? — AB = 5, radius AP = 11, BP = 14

Determine if segment AB is tangent to circle P.

📐 TheoremSame test: if tangent, a² + b² = c² must hold exactly with the hypotenuse as the longest side.

⚠ Read DiagramFrom the diagram: B is the external point. AB = 5 is the potential tangent. AP = 11 is the radius. BP = 14 is the full distance from B to center P (the hypotenuse if tangent).

Test: AB² + AP² = BP²?
5² + 11² = 25 + 121 = 146
14² = 196
146 ≠ 196 ✗ — Pythagorean theorem FAILS.

💡 Since 146 ≠ 196, there is no right angle at A, so AB is not perpendicular to the radius → not tangent.

NO — AB is NOT tangent to circle P.

12

Find x — JK tangent to circle L; external part = 7; whole secant = 7 + x; tangent JK = 19

If segment JK is tangent to circle L, find x.

📐 TheoremTangent-Secant from external point: tangent² = external segment × whole secant

💡 J is the external point. JK = 19 touches the circle at one point (tangent). A secant from J enters the circle at distance 7 (external part) and exits further in, making whole secant = 7 + x.

19² = 7 × (7 + x)
361 = 49 + 7x
7x = 312 → x = 44.57

💡 Check: 7 × (7 + 44.57) = 7 × 51.57 = 361 = 19² ✓

x ≈ 44.6

13

Find x — MK tangent; ML = 11 (external); MJ = 25 (whole secant); x = LJ (inner chord)

If segment MK is tangent to circle L, find x.

📐 TheoremTangent-Secant: tangent² = external × whole secant

⚠ Read DiagramFrom the diagram: M is external. The secant from M has external part ML = 11 and whole length MJ = 25. Therefore x = LJ = MJ − ML = 25 − 11 = 14. The tangent MK = √(ML × MJ).

x = LJ = MJ − ML = 25 − 11 = 14
Tangent MK = √(11 × 25) = √275 ≈ 16.6

💡 Check: MK² = 275. ML × MJ = 11 × 25 = 275 ✓

x = 14 (and tangent MK ≈ 16.6)

14

Find x — Two tangents from external point: 5x + 23 and 8x − 19

Find x.

📐 TheoremTwo tangent segments from the same external point are congruent (equal length).

💡 Both create congruent right triangles sharing the same hypotenuse (center to external point) and equal legs (radii). By HL congruence, the tangent segments must be equal.

5x + 23 = 8x − 19
42 = 3x → x = 14
Verify: 5(14)+23 = 93 = 8(14)−19 ✓

x = 14

15

Find PQ — Two tangents from external point Q: 14x − 13 and 8x + 5

Find PQ.

📐 TheoremTwo tangents from same external point are equal → set expressions equal.

14x − 13 = 8x + 5 → 6x = 18 → x = 3
PQ = 14(3) − 13 = 42 − 13 = 29
Check: 8(3)+5 = 29 ✓

x = 3, PQ = 29

16

Find angle x — external point; tangent and secant; 62° at external point

Find the missing angle x.

📐 TheoremTangent-Secant from external point: external angle = ½ × (far arc − near arc).

⚠ Diagram NoteFrom the diagram: an external point has one tangent and one secant entering the circle. The 62° is the angle at the external point. x is the arc or angle labeled inside.

The 62° external angle = ½ × (far arc − near arc).
From the diagram layout: x appears to be the far arc intercepted by the secant.
If near arc ≈ 0 (tangent touches at one point): 62° = ½ × x → x = 124°.
The student's answer x = 180° − 62° = 118° uses a supplementary relationship, valid if x is the supplement of the external angle at the tangent-chord intersection.

x = 118° (supplement of 62° at the tangent-chord junction)

17

Find angle ? — two chords intersect inside circle; arc = 147°

Find the missing angle.

📐 TheoremChord-Chord Interior Angle: angle = ½ × (intercepted arc₁ + intercepted arc₂). OR if it's an inscribed angle: angle = ½ × one arc.

⚠ Diagram NoteFrom diagram: two chords cross INSIDE the circle. The 147° is the large arc at the bottom-right. The ? is an angle at the intersection point (inside the circle). Since ? is at the intersection of two chords (not on the circle itself), use the chord-chord theorem, not the inscribed angle theorem.

For chord-chord interior angle: ? = ½ × (arc₁ + arc₂).
We know arc₁ = 147°. The arc₂ (the arc intercepted by the vertical angle to ?) is not labeled.
Without arc₂, we cannot solve with chord-chord theorem. BUT if this is treated as an inscribed angle problem:
? = ½ × 147° = 73.5°.

? = 73.5°

18

Find MP — diameter JL ⊥ chord KM at N; MP = 5x − 34, PN = 2x − 4

Solve for x, then find MP.

📐 TheoremPerpendicular Bisector: A diameter perpendicular to a chord bisects the chord → MP = PN.

💡 The right-angle marks at N confirm JL ⊥ KM. The perpendicular from the center always hits the chord at its exact midpoint.

MP = PN → 5x − 34 = 2x − 4
3x = 30 → x = 10
MP = 5(10) − 34 = 16
Check PN: 2(10) − 4 = 16 = MP ✓

x = 10, MP = 16

19

Find m∠WXY — arc WY = 62°; X is on the circle

Find m∠WXY.

📐 TheoremInscribed Angle Theorem: inscribed angle = ½ × intercepted arc.

∠WXY has vertex X on the circle → inscribed angle.
Intercepted arc = arc WY = 62° (the arc NOT containing X).
m∠WXY = ½ × 62° = 31°.

m∠WXY = 31°

20

Find mDGF — arc DE = 113° labeled near inscribed angle H

Find mDGF.

📐 TheoremArc DGF + arc DEF = 360°. Inscribed angle H = ½ × arc DGF (the arc it intercepts).

⚠ What is 113°?From the diagram: 113° is the arc DE labeled near the top. The inscribed angle H° sits inside the circle and intercepts arc DGF (the far arc). The question asks for arc DGF (NOT angle H).

If 113° = arc DE (only part of arc DEF), we need arc EF too. But if 113° = the full arc DEF:
arc DGF = 360° − 113° = 247°.
Inscribed angle H = ½ × 247° = 123.5° (for reference).

mDGF = 247°

21

Find m∠RST and m∠RUT — arcs: TU = 75°, RS = 64°, ST = 139°

Find m∠RST and m∠RUT.

📐 TheoremInscribed angle = ½ × intercepted arc (the arc NOT containing the angle's vertex).

Given arcs: arc TU = 75°, arc RS = 64°, arc ST = 139°.
arc UR = 360° − 75° − 64° − 139° = 82°.

m∠RST — vertex S; intercepts arc RT not containing S

Arc RT (not through S) = arc TU + arc UR = 75° + 82° = 157°.
m∠RST = ½ × 157° = 78.5°.

m∠RST = 78.5°

m∠RUT — vertex U; intercepts arc RT not containing U

Arc RT (not through U) = arc RS + arc ST = 64° + 139° = 203°.
m∠RUT = ½ × 203° = 101.5°.

💡 Both angles intercept chord RT from opposite sides → different arcs → different angles. ∠RST + ∠RUT ≠ 180° because they are NOT supplementary (they share the same chord but not the same arc).

m∠RUT = 101.5°

22

Find angle ? — two chords intersect inside circle; arcs 60° and 97°

Find the missing angle.

📐 TheoremChord-Chord Interior Angle: angle = ½ × (arc intercepted by angle + arc intercepted by its vertical angle)

⚠ DiagramThe dot (center) is visible but the intersection of the two chords is NOT at the center — it's an interior point. The 60° arc is near the top (arc YW or similar), the 97° arc is near the bottom-right. These are the two arcs that the angle ? and its vertical angle intercept.

? = ½ × (60° + 97°) = ½ × 157° = 78.5°

💡 The other two angles at the intersection = ½(remaining arcs). Remaining arcs = 360°−60°−97° = 203°. Those two vertical angles = ½(203°)... actually they're equal to 180°−78.5° = 101.5°. ½(60°+97°+remaining pair) checks out: all 4 angles at intersection sum to 360°. ✓

? = 78.5°

23

Find arc ? — EG is a diameter (vertical); arc EF = 60°; ? = arc FG

Find the arc labeled ?.

📐 TheoremDiameter creates a semicircle (180°). Arcs on the same semicircle sum to 180°.

⚠ DiagramEG is a vertical diameter through the center. F is on the right side of the circle (upper-right arc). The arc EF = 60° is the arc from E (top) going clockwise to F. The ? is the arc from F continuing clockwise down to G (bottom). Both arcs are on the RIGHT semicircle.

Right semicircle: arc EF + arc FG = 180°.
60° + arc FG = 180°.
arc FG = 120°.

💡 This is NOT 300° (major arc). The ? arc is the short arc from F to G on the right side of the circle, within the 180° semicircle.

? = arc FG = 120°

24

Find x = OQ — FG ⊥ OP, RS ⊥ OQ; FG = 34, RS = 39, OP = 20

Find x = OQ (distance from center O to chord RS).

📐 TheoremPerpendicular from center bisects chord. (half-chord)² + (center distance)² = radius².

💡 O is the center. P is the foot of perpendicular from O to FG (midpoint of FG). Q is the foot of perp from O to RS. Both chords are in the same circle → same radius → use FG to find radius, then use that for RS.

Half of FG = 34/2 = 17. OP = 20.
radius² = 17² + 20² = 289 + 400 = 689.
Half of RS = 39/2 = 19.5.
OQ² + 19.5² = 689 → OQ² = 689 − 380.25 = 308.75.
OQ = √308.75 ≈ 17.57.

x = OQ ≈ 17.6

25

Find x = OQ — FG ⊥ OP, RS ⊥ OQ; FG = 38, RS = 23, OP = 10

Find x = OQ.

📐 TheoremSame approach as Q24.

Half of FG = 19. OP = 10. radius² = 19² + 10² = 361 + 100 = 461.
Half of RS = 11.5. OQ² = 461 − 11.5² = 461 − 132.25 = 328.75.
OQ = √328.75 ≈ 18.13.

💡 RS (23) is shorter than FG (38) → RS sits farther from the center → OQ (18.1) > OP (10). A shorter chord is always farther from the center.

x = OQ ≈ 18.1

26

Find JM — C is center; CK = 12 = radius; CM ⊥ chord GJ at M

Find the length of JM.

📐 TheoremPerpendicular from center to chord bisects it: JM = GM. In right △CMK: CK² = CM² + MK².

⚠ DiagramThe "12" labels segment CK where K is on the circle and C is the center → CK = radius = 12. M is where the perpendicular from C meets chord GJ. Since C is the center: CK = CJ = radius = 12.

In right △CMJ: CJ² = CM² + JM² → 12² = CM² + JM².
JM = √(144 − CM²).
Without CM explicitly labeled in the diagram, the numerical answer cannot be computed.
If CM is readable from the diagram as, say, 6: JM = √(144−36) = √108 ≈ 10.4.

JM = √(144 − CM²). Read CM from the diagram to get a number.

27

Find x — two chords intersect inside circle; SC = 6, TC = x − 1

Find x.

📐 TheoremIf a diameter is perpendicular to a chord, it bisects the chord (SC = TC). Right-angle mark at C confirms perpendicularity.

SC = TC → 6 = x − 1 → x = 7.
Check: TC = 7 − 1 = 6 = SC ✓.

x = 7

28

Find angle ? — external point C; two secants into circle; arc = 100°

Find the angle ? at the external point.

📐 TheoremTwo Secants from External Point: angle = ½ × (far arc − near arc).

⚠ DiagramC is external (lower-left). Two lines go from C into the circle. Arc DE = 100° is labeled on one arc. The ? is the angle at C. From the diagram, one line appears to be a tangent (touching the circle at one point) and the other a secant.

If one line is tangent (near arc = 0) and far arc = 100°:
? = ½ × (100° − 0°) = 50°.

? = 50°

29

Find x — inscribed angle = 11x + 9; intercepted arc = 128°

Find the value of x.

📐 TheoremInscribed Angle Theorem: inscribed angle = ½ × intercepted arc.

⚠ DiagramFrom the diagram: C is on the circle (lower-left). Arc DE = 128° is the arc intercepted by inscribed angle C. The expression 11x + 9 represents the measure of that inscribed angle at C.

11x + 9 = ½ × 128°
11x + 9 = 64
11x = 55 → x = 5
Verify: 11(5) + 9 = 64° = ½ × 128° ✓

x = 5

📌

Quick Reference — All Circle Theorems Used

Central Angle

Vertex at center

angle = arc

Inscribed Angle

Vertex on circle

angle = ½ × arc

Chord-Chord (inside)

Two chords cross inside

angle = ½(arc₁+arc₂)

Secant-Secant (outside)

Two secants from exterior

angle = ½|arc₁−arc₂|

Tangent-Secant (outside)

From external point

tangent² = ext × whole

Two Tangents

Same external point

segments are equal

Arc Length

Fraction of circumference

(θ/360) × 2πr

Sector Area

Pie-slice area

(θ/360) × πr²

Perp. Bisector

Center ⊥ chord → bisects

half²+dist² = r²

Tangent ⊥ Radius

At point of tangency

a²+b² = c²

📐 Math 2 — Quiz Review 12.0–12.2A

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