Circles (వృత్తములు)

r = 7 cm

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

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.

Chord ని bisect చేస్తుంది (Perpendicular Bisector Theorem)

Theorem: Center నుండి chord కి perpendicular వేస్తే chord ని exactly midpoint వద్ద bisect చేస్తుంది.
OM ⊥ AB → AM = MB.

Converse: Chord midpoint ను center తో కలిపితే chord కి perpendicular అవుతుంది.
చాలా problems లో ఈ property use అవుతుంది!

Same arc పై (same segment లో) inscribed angles సమానంగా ఉంటాయి

Theorem: Same chord AB యొక్క same side లో ఉన్న circle పై అన్ని inscribed angles equal.
∠APB = ∠AQB (P, Q same segment లో ఉంటే).

ఉదా: ఒక stadium చుట్టూ ఉన్న అన్ని spectators కి same chord "football goal" same angle లో కనిపిస్తుంది!

Common chord పై ప్రతి point నుండి రెండు circles కి equal power

Common chord AB పై point P:
Power w.r.t. circle 1: PA×PB (chord P వద్ద).
Power w.r.t. circle 2: PA×PB (same, since A,B on both circles).
Equal power → P on radical axis.
కాబట్టి common chord ⊆ radical axis.
Both perpendicular to line of centers → they coincide!

PC = 9

PA² = PB × PC.
36 = 4 × PC → PC = 9 cm.

24 cm

AM² = r² - OM² = 169-25 = 144 → AM=12.
Chord = 2×AM = 24 cm.

70°

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

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.

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!

PA = 12 cm

OA⊥PA → PA²=OP²-OA²=169-25=144 → PA=12.
Also PB=12 (equal tangents).

66 cm

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

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).

(x-3)²+(y+4)²=25

Center (3,-4), r=5:
(x-3)²+(y-(-4))²=5².
(x-3)²+(y+4)²=25.

Expanded: x²-6x+9+y²+8y+16=25 → x²+y²-6x+8y=0.

∠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° ✓

C = 2πr (లేదా πd)

Circumference = circle యొక్క perimeter (boundary length).
C = 2πr = πd.
π ≈ 3.14159... (irrational number).

ఉదా: r=7 → C = 2×(22/7)×7 = 44 cm.
గుర్తు: π = 22/7 approximation exam లో చాలా use అవుతుంది!

64π cm²

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

OT = 5 cm

OT ⊥ PT → OT²+PT²=OP².
OT²+144=169.
OT²=25 → OT=5 cm.

5-12-13 Pythagorean triple! గుర్తు: tangent problems లో Pythagoras use చేయాలి.

220°

Inscribed ∠B=70° → minor arc AC = 2×70° = 140°.
Major arc AC = 360°-140° = 220°.

2:1

Central angle : Inscribed angle = 2:1.
Same arc పై central = 2× inscribed.

ఉదా: Arc AB subtends 80° at center → 40° at any point on remaining circle.

r = 7 cm

r²+24²=25².
r²=625-576=49 → r=7 cm.

7-24-25 Pythagorean triple!

ఒకే center ఉన్న వేర్వేరు radius గల circles

Concentric circles = same center కానీ different radii.
ఉదా: Target board (archery) — అన్ని circles same center.
Ripples in water = concentric circles!

గుర్తు: "Con" = together, "centric" = center — same center circles.

50°

PA=PB (equal tangents) → △PAB isosceles.
∠PAB=∠PBA=(180°-80°)/2=100°/2=50°.

PA × PB = PC × PD

Intersecting Chords Theorem:
Circle లో రెండు chords P వద్ద intersect అయినప్పుడు:
PA × PB = PC × PD.

Proof: △PAC ~ △PDB (AA similarity) → PA/PD = PC/PB → PA×PB = PC×PD.
ఉదా: PA=3, PB=8, PC=4 → PD=6. 3×8=24=4×6 ✓

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!

Normal: y/x = y₁/x₁ లేదా xy₁ = yx₁

Normal at P(x₁,y₁) to circle x²+y²=r²:
Normal passes through center (0,0) and P(x₁,y₁).
Equation: y/x = y₁/x₁ → xy₁-yx₁=0.

Normal always passes through center of circle!
This is because radius ⊥ tangent at P, and normal = radius direction.

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 ✓)!

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!

AC×BD = AB×CD + BC×DA = 15+24 = 39

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

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!

Area=50π; Perimeter=πr+2r=(π+2)×10

r²=100 → r=10.
Semicircle area = 100π/2 = 50π.
Semicircle perimeter = πr+2r = 10π+20 = 10(π+2).

120°

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

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

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!

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!

Center=(-D/2, -E/2); r=½√(D²+E²-4F)

x²+y²+Dx+Ey+F=0 completing the square:
(x+D/2)²+(y+E/2)² = D²/4+E²/4-F.
Center = (-D/2, -E/2).
r = ½√(D²+E²-4F).

Condition for real circle: D²+E²-4F > 0.

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 ✓.

Triangle circumcircle పై point P నుండి మూడు sides కి feet collinear అయినప్పుడు ఆ line Simson line

Simson Line Theorem:
Triangle ABC circumcircle పై point P నుండి BC, CA, AB కి perpendiculars వేస్తే feet D, E, F collinear → DEF = Simson line.

Converse: D,E,F collinear → P circumcircle పై ఉంటుంది.
Elegant theorem connecting circles and lines!

r = 7 cm

2πr = 44.
2×(22/7)×r = 44.
r = 44×7/(2×22) = 308/44 = 7 cm.

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.

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!

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!

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.

x²+y²=16 (Circle center origin, r=4)

PA²=(x-3)²+y², PB²=(x+3)²+y².
PA²+PB²=2x²+18+2y²=50.
2(x²+y²)=32.
x²+y²=16. Circle, r=4!

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!

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!

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)!

3 side midpoints + 3 altitude feet + 3 Euler points

Nine-point circle:
1. మూడు sides midpoints.
2. మూడు altitudes foot points.
3. Orthocenter నుండి మూడు vertices midpoints (Euler points).

Nine-point circle radius = circumradius/2.
Feuerbach theorem: Nine-point circle tangent to incircle!

8 cm

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

3-4-5 Pythagorean triple!

40°

Alternate Segment Theorem:
Tangent-chord angle = inscribed angle in alternate segment.
∠QPT=40° → angle in alternate segment = 40°.

sin A = 3/5

Sine rule: a/(sin A) = 2R.
12/sin A = 20 → sin A = 12/20 = 3/5.

Area = ½bc×sinA. Also Area=abc/(4R).
24=12×bc/(40) → bc=80.
And ½×bc×sinA=24 → bc×3/5=48 → bc=80 ✓.