Circles — Tangents & Theorems 🔵

Tangent, secant, do important theorems — aur unke proofs Roman Hindi mein.

🔵 Circle — Important Terms

🎡 Ferris Wheel se samjho:

Ek Ferris wheel ek circle hai. Jab ek car wheel zameen ko chhooti hai — woh point tangent point hai. Zameen line tangent hai. Ek chotti si touch, phir alag — yahi tangent ka matlab hai!

O r = radius diameter chord arc ⌢ tangent

Tangent

Ek line jo circle ko exactly ek point par chhooti hai (touch point = point of tangency)

Secant

Ek line jo circle ko do points par cut karti hai

Chord

Circle ke do points ko milane wali line segment

Arc

Circle ka koi bhi hissa (curved part)

📏 Theorem 1: Tangent ⊥ Radius

Theorem: Tangent, point of contact par radius se perpendicular hoti hai.
Matlab: agar TP tangent hai aur O center, to OT ⊥ TP (∠OTP = 90°)
Proof by Contradiction
Maano OP ⊥ TP nahi hai. To ek aur point Q exist karta hai TP par jaise OQ ⊥ TP.
To OQ < OP (kyunki perpendicular distance sabse choti hoti hai).
Lekin Q TP par hai aur TP tangent hai — TP circle ko sirf T par chhooti hai.
Q circle ke bahar hoga → OQ > r.
Lekin OT = r (radius) aur OQ < OT (humari assumption) → OQ < r.
Contradiction! Q circle ke andar nahi ho sakta (tangent se circle cut nahi hoti).
∴ OP hi perpendicular hai → OT ⊥ TP ■

📏 Theorem 2: Equal Tangents from External Point

Theorem: Ek external point se kheenchi gayi do tangent lines ki length barabar hoti hai.
Agar PA aur PB circle ke tangents hain external point P se, to PA = PB
P A B O PA PB PA = PB !
Proof by RHS Congruence
In △OAP aur △OBP:
OA = OB (radii of same circle)
∠OAP = ∠OBP = 90° (tangent ⊥ radius)
OP = OP (common)
By RHS: △OAP ≅ △OBP
∴ PA = PB (CPCT) ■
Important corollary:
∠POA = ∠POB (OP bisects ∠AOB)
∠APO = ∠BPO (OP bisects ∠APB)
OP ye dono angles bisect karta hai!

📐 Number of Tangents from a Point

Point PositionNumber of TangentsDiagram
Inside circle (interior)0No tangent possible
On circle (boundary)1 (exactly)One tangent at that point
Outside circle (exterior)2Two tangents PA and PB

🔢 Important Angle Results

Example 1: PA aur PB tangents, ∠APB = 80°. ∠AOB nikalo.
Quad AOBP mein: ∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + 80° + ∠AOB = 360°
∠AOB = 100°
Example 2: ∠APB = 60°. ∠OAB nikalo.
PA = PB → △APB isosceles → ∠PAB = ∠PBA = (180°−60°)/2 = 60°
∠OAP = 90° (tangent ⊥ radius)
∠OAB = ∠OAP − ∠PAB = 90° − 60° = 30°
Example 3: Do circles internally touch karte hain. Common tangent ki length? (r₁=5, r₂=3, d=2)
Internal tangent length: agar internally tangent hain, common internal tangent 0 hai.
Direct common tangent length = √(d²−(r₁−r₂)²) = √(4−4) = 0. Confirms internal tangency.
(Ye special case — ek hi common tangent exist karta hai)

📐 Tangent from External Point — Length

Tangent Length Formula
Agar P external point hai, O center, r radius:
Tangent length = √(OP² − r²)

(Pythagoras theorem: OP² = OA² + PA² → PA = √(OP²−r²))
Example: Center O, radius 5 cm. External point P, OP = 13 cm. Tangent length?
PA = √(13² − 5²) = √(169−25) = √144 = 12 cm

🔄 Circle aur Quadrilateral

Theorem: Ek quadrilateral ke agar sare 4 sides ek circle (inscribed circle) ko tangent karte hain, to:
AB + CD = BC + AD
Example: Quadrilateral ABCD mein AB=6, BC=7, CD=4. AD nikalo (circle inscribed).
AB + CD = BC + AD
6 + 4 = 7 + AD → AD = 3 cm

📝 Practice Questions

Q1. TP tangent hai, TO = 10 cm, OP = 26 cm. TP nikalo.

TP = √(OP²−TO²) = √(676−100) = √576 = 24 cm

Q2. PA aur PB tangents, ∠AOB = 120°. ∠APB nikalo.

∠APB = 180° − ∠AOB = 180° − 120° = 60°

Q3. Prove karo ki tangent point par circle aur tangent ke beech angle = chord ke opposite arc mein inscribed angle.

Yeh Tangent-Chord Angle theorem hai. ∠between tangent and chord = inscribed angle in alternate segment. Use: ∠TAB = ∠ACB (alternate segment theorem). Proof: ∠OAT=90°, ∠OAB = 90°−∠TAB. In △OAB (isosceles, OA=OB=r), ∠OAB=∠OBA → ∠AOB=180°−2∠OAB. Inscribed angle ∠ACB = ∠AOB/2 = 90°−∠OAB = ∠TAB ■

Q4. Rhombus ABCD ek circle ke bahar hai. Dikhao ki AB = BC. (Circle sabhi 4 sides ko tangent karta hai)

Rhombus mein AB=BC=CD=DA (definition). Tangent condition bhi satisfy hoti hai (AB+CD=BC+DA → 2AB=2AB ✓). Yeh ek special case hai jab rhombus ka inscribed circle exist karta hai.
🎮 Geometry Lab →