Chord of Larger Circle Touching Smaller Circle
Two concentric circles have the same center O with radii $R > r$. A chord $AB$ of the larger circle touches the smaller circle at point $P$.
Chord AB of C₁ touches C₂ at P
Therefore, Chord AB = 2 × AP = 2 × √(R² - r²)
Essential Exam Proofs
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC
Let circle touch sides AB, BC, CD, DA at points P, Q, R, S respectively.
By Theorem 10.2 (Lengths of external tangents are equal):
BP = BQ --- (Eq 2)
CR = CQ --- (Eq 3)
DR = DS --- (Eq 4)
Adding equations (1) + (2) + (3) + (4):
AB + CD = AD + BC (Proved)
Prove that tangents drawn at the ends of a diameter of a circle are parallel
Let AB be a diameter of a circle with center O. Let lines PQ and RS be tangents at points A and B respectively.
By Theorem 10.1 (Radius ⊥ Tangent):
OB ⊥ RS ⟹ ∠OBS = 90°
∠OAQ = ∠OBS = 90°
Since alternate interior angles are equal (∠OAQ = ∠OBS), the lines are PQ ∥ RS (Proved).
Key Takeaways
Concentric circles chord touching inner circle is bisected at point of contact.
Circumscribed quadrilateral property: Sum of opposite sides are equal (AB + CD = AD + BC).
Tangents at ends of a diameter are parallel (alternate interior angles = 90°).