Circles (వృత్తములు)

AC×BD = AB×CD + BC×DA = 15+24 = 39

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

r = 10 cm

Perpendicular from center bisects chord.
AM = 8 cm (half of chord).
OA² = OM² + AM² = 36+64 = 100.
r = OA = 10 cm.

8 cm

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

3-4-5 Pythagorean triple!

Tangent-chord angle = inscribed angle in alternate segment

Theorem (Alternate Segment Theorem / Tangent-Chord Angle):
Tangent PQ మరియు chord AB వద్ద angle = alternate segment లో inscribed angle.

∠PAB = ∠ACB (C alternate segment లో).
ఇది "Tangent-chord angle = angle in alternate segment"!
Exam లో చాలా important theorem.

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!

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!

8 cm

Externally tangent: d = r₁ + r₂ = 5+3 = 8 cm.
Internally tangent: d = |r₁-r₂| = |5-3| = 2 cm.

గుర్తు:
External tangency → centers far apart → d = r₁+r₂.
Internal tangency → one inside other → d = r₁-r₂.

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.

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).

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.

(k₁+k₂+k₃+k₄)² = 2(k₁²+k₂²+k₃²+k₄²)

Descartes' Circle Theorem:
Curvature k = 1/r.
(k₁+k₂+k₃+k₄)² = 2(k₁²+k₂²+k₃²+k₄²).

Used to find the 4th circle tangent to 3 given mutually tangent circles.
Apollonian gasket creates infinite tangent circles from this theorem!

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 ✓.

Center=(2,3), r=2

Complete square:
(x²-4x+4)+(y²-6y+9) = -9+4+9 = 4.
(x-2)²+(y-3)²=4.
Center=(2,3), r=2.

r = 7 cm

r²+24²=25².
r²=625-576=49 → r=7 cm.

7-24-25 Pythagorean triple!

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. ✓

Chord మరియు arc మధ్య ఉన్న region

Segment = chord తో cut అయిన circle part.
Minor segment = smaller part (minor arc + chord).
Major segment = larger part.

Sector vs Segment:
Sector = Pie slice (2 radii + arc).
Segment = Chord + arc (radii లేదు).
గుర్తు: Segment లో center ఉండదు (minor segment లో)!

4 common tangents

Two circles external గా (d > r₁+r₂) అయినప్పుడు 4 common tangents.
d=10, r₁+r₂=8. 10>8 → External circles.
4 tangents: 2 direct + 2 transverse.

Cases:
d > r₁+r₂ → 4 tangents (external)
d = r₁+r₂ → 3 tangents (externally tangent)
r₁-r₂ < d < r₁+r₂ → 2 tangents (intersecting) d = r₁-r₂ → 1 tangent (internally tangent) d < r₁-r₂ → 0 tangents (one inside other)

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.

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!

xx₁ + yy₁ = r²

Chord of contact (line joining two tangent points):
xx₁ + yy₁ = r².

Same form as tangent at a point!
Tangent at (x₁,y₁): xx₁+yy₁=r².
Chord of contact from P(x₁,y₁): xx₁+yy₁=r².

But: Tangent → point ON circle. Chord of contact → point OUTSIDE circle.

l² + d² = r²

Perpendicular bisects chord.
Half chord = l, radius = OA = r, distance = d = OM.
Right triangle OMA: l²+d²=r².

ఇది standard relation: l=√(r²-d²), d=√(r²-l²).

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.

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!

6√2 cm

Square inscribed in circle: diagonal = diameter = 12.
Side = diagonal/√2 = 12/√2 = 6√2 cm.

Area of square = side² = (6√2)² = 72 cm².

220°

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

2:1

Area ∝ r². Area ratio = 4:1 = r₁²:r₂².
r₁/r₂ = √4 = 2. Ratio = 2:1.

40°

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

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!

r = 7 cm

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

అవుతుంది (3²+4²=25); Tangent: 3x+4y=25

3²+4²=9+16=25 ✓ → point on circle.
Tangent at (x₁,y₁) to x²+y²=r²: xx₁+yy₁=r².
Tangent at (3,4): 3x+4y=25.

Standard formula: Tangent at (x₁,y₁) to circle x²+y²=r² is xx₁+yy₁=r².

50°

∠AOB = 2×∠ACB = 80° (central=2×inscribed).
OA=OB (radii) → △OAB isosceles.
∠OAB = ∠OBA = (180°-80°)/2 = 50°.

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!

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!

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.

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).

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 ✓)!

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!

OS = 9 cm

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

66 cm

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

OM = 5 cm

AM = 12 (half chord).
OM² = OA²-AM² = 169-144=25 → OM=5 cm.

5-12-13 triple!

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.

PA = 12 cm

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

50°

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

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!

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 ✓

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.

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.

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.