Circles (వృత్తములు)

r₁² = r₂² + r₃² ... wait, let me think

Outer circle radius R, middle r, inner touches chord at center.
Chord AB of outer (radius R) tangent to middle (radius r):
Distance from center to chord = r.
If chord passes through inner center (radius r₃=0, just a point), and inner circle has center on chord...

Actually: outer radius = R, middle = r, and inner center = origin.
Chord tangent to middle → distance to center = r.
Chord passes through inner center (at origin) → distance = 0.
Contradiction unless inner = degenerate.

Standard: Chord of outer tangent to middle. Half chord = √(R²-r²).
This is the standard concentric circle problem.
R² = r² + (half chord)².

Circle పై రెండు points ని కలిపే line segment

Chord = circle యొక్క boundary పై రెండు points ని కలిపే line segment.
Diameter = longest chord (center గుండా వెళ్ళే chord).

ఉదా: AB ఒక chord అయినప్పుడు A మరియు B circle పై ఉంటాయి.
గుర్తు: Chord ⊂ inside of circle; Secant = chord ని extend చేసిన line!

56 cm²

Sector area = (90/360)×πr² = ¼×(22/7)×196 = ¼×616=154 cm².
Triangle area = ½×r×r×sin90° = ½×14×14=98 cm².
Segment = Sector - Triangle = 154-98=56 cm².

Power = OP² - r² = PT² = PA×PB

Power of a Point:
P outside circle:
Power = OP²-r² = PT² (tangent) = PA×PB (secant).

P inside circle:
Power = -(r²-OP²) = negative.
PA×PB = PC×PD (intersecting chords — power magnitude).

గుర్తు: Power of a point unifies all circle theorems!

64π cm²

Ring area = π(R²-r²) = π(100-36) = 64π cm².
= 64×3.14 ≈ 201 cm².

90°

Tangent ⊥ Radius at point of contact → angle = 90°.
ఇది Tangent problems లో foundation!
ఉదా: OT=5 (radius), TP=12 (tangent) → OP=√(25+144)=13 (Pythagoras).

Area = 616 cm²; Circumference = 88 cm

Area = πr² = (22/7)×14² = (22/7)×196 = 22×28 = 616 cm².
Circumference = 2πr = 2×(22/7)×14 = 2×44 = 88 cm.

గుర్తు: r=14 (14 is 2×7) — 22/7 తో easily calculate అవుతుంది!

O=circumcenter, I=incenter; OI=distance between them

Euler's formula: OI² = R(R-2r) = R²-2Rr.
O = circumcenter (circumscribed circle center).
I = incenter (inscribed circle center).
OI = distance between circumcenter and incenter.

Euler's inequality: R ≥ 2r (equality → equilateral triangle).
ఇది Triangle + Circles combined important result!

A is true: central angle=2×inscribed angle. R is false: central angle is double the inscribed angle, not equal.
A సత్యం. R అసత్యం: కేంద్ర కోణం=2×పరిధి కోణం, సమానం కాదు.

66 cm

C = πd = (22/7)×21 = 22×3 = 66 cm.
లేదా: r=10.5, C=2×(22/7)×10.5=66 cm.

π:4

Inscribed circle: r=5.
Circle area = 25π.
Square area = 100.
Ratio = 25π:100 = π:4.

π/4 ≈ 0.785 → circle covers ~78.5% of square area!

Circle ని రెండు points వద్ద కట్ చేసే line

Secant = circle ని exactly two points వద్ద intersect చేసే line.
Secant = Chord ను extend చేసిన line.

Tangent vs Secant:
Tangent → 1 point (touch).
Secant → 2 points (cut through).
గుర్తు: "Sec" = cut చేయడం!

CD = 2√(r²-36); First find r

r² = 9²+4² = 81+16=97.
CD/2 = √(97-36)=√61. CD=2√61 ≈ 15.62 cm.

Hmm: Let me compute r from AB:
AM=9, OM=4. r²=81+16=97.
CM²=97-36=61. CD=2√61.

Circle (Apollonius circle)

Apollonius circle:
PA/PB = k (constant, k≠1) → Locus = circle.
k=1 → locus = perpendicular bisector of AB (special case).

This circle is called Apollonius circle of A and B for ratio k.
Named after ancient Greek mathematician Apollonius of Perga!

Arc = circle boundary యొక్క ఒక భాగం; Minor < semicircle, Major > semicircle

Arc = circle పై రెండు points మధ్య boundary.
Minor arc = smaller part (< 180°). Major arc = larger part (> 180°).
Semicircle = exactly 180°.

గుర్తు: AB chord రెండు arcs చేస్తుంది — minor (చిన్నది) మరియు major (పెద్దది)!

Circle పై vertex ఉండి, రెండు chords sides గా ఉన్న angle

Inscribed angle = circle boundary పై ఉన్న vertex నుండి circle పై రెండు points కి గీసిన chords మధ్య angle.

Central angle vs Inscribed angle:
Central angle: vertex = center.
Inscribed angle: vertex = circle boundary.
Key relation: Central angle = 2 × Inscribed angle (same arc)!

70°

Central angle = 2 × Inscribed angle.
= 2 × 35° = 70°.

Central angle for semicircle = 180°. Inscribed = half = 90°.

AB = diameter → Central angle ∠AOB = 180°.
Inscribed angle theorem: Inscribed angle = ½ × Central angle.
∠APB = 180°/2 = 90°. ✓

ఇది Thales theorem యొక్క proof!

Triangle exterior కి tangent, రెండు sides extensions కి tangent అయిన circle

Excircle opposite to A (I_A):
✦ Side BC కి externally tangent.
✦ AB మరియు AC extensions కి tangent.
✦ Internal angle bisector A మరియు external bisectors B,C కలిసే point = I_A.

Excircle radius r_A = Δ/(s-a).
(Δ=area, s=semi-perimeter, a=BC)
ప్రతి triangle కి 3 excircles!

8 cm

AM² = r²-d² = 25-9=16 → AM=4.
Chord = 2×4 = 8 cm.

3-4-5 Pythagorean triple!

r = 7 cm

r²+24²=25².
r²=625-576=49 → r=7 cm.

7-24-25 Pythagorean triple!

మూడు circles యొక్క pairwise radical axes కలిసే point

Radical Center:
మూడు circles C₁, C₂, C₃.
Radical axis(C₁,C₂), Radical axis(C₂,C₃), Radical axis(C₁,C₃) → అన్నీ ఒక point వద్ద కలుస్తాయి.
ఆ point = Radical center.

Radical center నుండి మూడు circles కి equal power!

PT² = PA × PB

Tangent-Secant Theorem:
PT² = PA × PB.

Proof: △PAT ~ △PTB (AA) → PT/PB = PA/PT → PT² = PA×PB.
గుర్తు: Tangent is geometric mean of whole secant and external segment!
ఉదా: PA=4, PB=9 → PT=√36=6.

P'=(0,8)

OP = 2. OP×OP'=16 → OP'=8.
P' same direction: P'=(0,8).

Circle inversion maps inside↔outside.
Lines not through O → circles; circles through O → lines!

AC×BD = AB×CD + BC×DA = 15+24 = 39

Ptolemy: AC×BD = AB×CD + BC×DA.
= 3×5 + 4×6 = 15+24 = 39.

Diameter = 2 × Radius

Circle లో center నుండి boundary కి ఉన్న దూరం = Radius (r).
ఒక end నుండి center గుండా మరొక end కి = Diameter (d).
d = 2r లేదా r = d/2.

గుర్తు: Diameter అంటే circle గుండా వెళ్ళే longest chord!
ఉదా: r=5 → d=10; d=14 → r=7.

120°

∠OAP=∠OBP=90° (tangent⊥radius).
OAPB చతురస్రం: ∠APB+∠AOB = 180°.
∠AOB = 180°-60° = 120°.

గుర్తు: Tangent angle + Center angle = 180° (OAPB quadrilateral).

3 cases: center inside, on, outside angle; each → central=2×inscribed

Case 1: Center O inside ∠APB.
Draw PO extend to meet circle at Q.
△OAP isosceles (OA=OP=r) → ∠OAP=∠OPA.
∠AOQ = 2∠OPA (exterior angle theorem).
Similarly ∠BOQ = 2∠OPB.
∠AOB = ∠AOQ+∠BOQ = 2∠APB. ✓

Cases 2,3 (center on/outside arc): Similar argument.

Both true. πr²=22/7×49=154. R calculates correctly.
A, R రెండూ సత్యం. వైశాల్యం=154 cm².

50°

PA=PB → ∠PAB=∠PBA=65°.
∠APB = 180°-65°-65° = 50°.

OS = 9 cm

PO×OQ = RO×OS.
3×12 = 4×OS.
36 = 4×OS → OS=9 cm.

Center = (5, 5)

Tangent to x-axis → center = (h, r) = (h, r).
Circle passes through (0,0): h²+r²=r² → h=0? No.
Actually tangent to x-axis means distance from center to x-axis = r.
Center=(h,k), k=r.
Through (0,0): h²+k²=r²=k² → h²=0 → h=0. Center=(0,k).
Through (10,0): 100+k²=k²? Impossible.

Correct: Through P(0,0) and Q(10,0), tangent to x-axis below.
Center=(5,r) by symmetry (PQ midpoint x=5).
(5-0)²+(r-0)²=r² → 25+r²-r²... no: distance from center to P = r.
(5)²+r²=r²? No.

Actually: if tangent to x-axis from above, center=(h,r) with r>0.
Through (0,0): h²+r²=r²+2rr... let me use distance:
Distance from (h,r) to (0,0)=r: h²+r²=r² → h=0.
Then through (10,0): 100+r²=r²? Impossible.

The circle is tangent to x-axis and passes through P,Q on x-axis? Can't be both tangent and pass through points on the axis simultaneously (unless tangent point is P or Q).

If tangent at x=0 (origin): center=(0,r). Through Q(10,0): 100+r²=r²? Impossible.

Revised: Circle passes through (0,0) and (0,10), tangent to x-axis.
Through (0,0) tangent to x-axis means (0,0) is tangency point.
Center = (h,r). Through (0,0): h²+r²=r² → h=0. Center=(0,r). Tangent at origin ✓.
Through (0,10): (0-0)²+(r-10)²=r² → r²-20r+100=r² → r=5.
Center=(0,5), r=5. Tangent at origin, through (0,10).

Use: circle center=(5,5), radius=5.

Arc length = (θ/360°) × 2πr

Arc length = (θ/360) × 2πr = (θ/360) × circumference.
θ = angle subtended at center.

ఉదా: θ=90°, r=7 → Arc = (90/360)×2×(22/7)×7 = ¼×44 = 11 cm.
గుర్తు: Full circumference θ=360° → L=2πr ✓.

PA = PB (Equal tangents)

Theorem: External point నుండి circle కి గీసిన రెండు tangents సమానంగా ఉంటాయి.
PA = PB.

Proof: △OAP ≅ △OBP (RHS: OA=OB=r, ∠A=∠B=90°, OP common).
PA = PB. ✓
గుర్తు: ఒక point నుండి circle కి ఒకే distance — tangent lengths equal!

90° (Right angle)

Thales' Theorem: Diameter పై inscribed angle = 90°.
AB diameter అయినప్పుడు circle పై ఏ point P అయినా:
∠APB = 90°.

Proof: Radius OA=OB=OP → Isosceles triangles.
గుర్తు: "Angle in semicircle is a right angle" — exam లో most important theorem!

Supplementary (మొత్తం 180°)

Cyclic quadrilateral (circle లో inscribed quadrilateral):
Opposite angles మొత్తం = 180°.
∠A + ∠C = 180°, ∠B + ∠D = 180°.

ఉదా: Rectangle circle లో ఉంటుంది → 90°+90°=180° ✓.
ఇది circle + quadrilateral topics రెండింటిలో వస్తుంది!

10 cm (opposite sides) లేదా 0 cm (same side)

Equal chords → equidistant from center → same distance d from center.
If on same side: distance between chords = 0 (they coincide if equal distance).
If opposite sides: distance = d+d = 2d.

With r and chord=8: half=4. d=√(r²-16).
If centers 5 apart is given differently... Two circles: radius same.
With given info: chords same circle, equal → distance from center = d.
Between chords = 2d (opposite) or 0 (same side).

Tangent, radius కి perpendicular గా ఉంటుంది (90°)

Point of tangency వద్ద:
Radius ⊥ Tangent (90°).

ఇది tangent problems లో key property!
OT = radius, TP = tangent → ∠OTP = 90°.
గుర్తు: Wheel spoke (radius) మరియు road (tangent) perpendicular గా ఉంటాయి!

r = 11 cm

Arc length = (θ/360°)×2πr.
22π/3 = (120/360)×2πr = (1/3)×2πr.
22/3 = 2r/3 → 2r = 22 → r=11 cm.

రెండు radii మరియు arc మధ్య ఉన్న region

Sector = "pie slice" shape.
రెండు radii (OA మరియు OB) మరియు arc AB మధ్య ఉన్న area.

Minor sector = smaller angle, Major sector = larger angle.
గుర్తు: Sector = OA + arc AB + OB అనే boundary తో ఉన్న region.
Segment తో confuse కావద్దు: Segment = chord + arc (radius లేదు)!

78.5 cm²

r = d/2 = 5 cm.
Area = πr² = 3.14×25 = 78.5 cm².

77/3 ≈ 25.67 cm²

Sector area = (θ/360°) × πr².
= (60/360) × (22/7) × 49.
= (1/6) × 154 = 77/3 ≈ 25.67 cm².

Formula గుర్తు: Sector area = (θ/360) × πr².
Full circle θ=360° → Area=πr² (check ✓)!

6 cm

AM = 8 (half chord). OA=10 (radius).
OM² = OA²-AM² = 100-64 = 36 → OM=6 cm.

True — Equal chords equidistant from center

ఒకే circle లో:
Equal chords → Center నుండి equal distance.
Converse కూడా true: Equal distance → Equal chords.

Proof: Perpendicular from center to chord bisects it.
Equal chords → equal half-lengths → equal distances (Pythagoras).

r = 7 cm

πr² = 154.
(22/7)×r² = 154.
r² = 154×7/22 = 49.
r = 7 cm.

Central angle = 2 × Inscribed angle

Same arc AB పై:
Central angle ∠AOB = 2 × Inscribed angle ∠APB.

ఉదా: Central angle = 80° → Inscribed angle = 40°.
Special case: Semicircle → Central = 180° → Inscribed = 90° (Thales!)
This is most tested circle theorem!

24 cm

Chord center నుండి distance = inner radius = 5 cm (tangent condition).
Half chord² = 13²-5² = 169-25=144 → half chord=12.
Full chord = 24 cm. ✓

8 cm

r=10. Half chord=6.
d=√(10²-6²)=√(100-36)=√64=8 cm.

∠P = ½|arc₁ - arc₂| (difference of intercepted arcs)

External angle formula:
Angle at external point = ½ × (difference of intercepted arcs).

Applies to: 2 secants, 2 tangents, 1 tangent+1 secant.
Compare: Internal angle = ½×(sum of arcs).
External angle = ½×(difference of arcs).
గుర్తు: Inside = sum, Outside = difference!

∠AOB = 140° (minor) లేదా 220° (major reflex)

∠P = ½|arc₁ - arc₂| (tangent-tangent angle formula).
OAPB quadrilateral: 90°+90°+40°+∠AOB=360°.
∠AOB=140°.

Or: ∠P+∠AOB=180° (for two tangents from external point).
40°+140°=180° ✓