Example-1 :-  Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

  Given that : Two concentric circles C1 and C2 with centre O and a chord AB of the larger 
               circle C1 which touches the smaller circle C2 at the point P . 
  Prove that : AP = BP.
  Construction : Let us join OP.
  Proof : Then, AB is a tangent to C2 at P and OP is its radius. Therefore, by Theorem 10.1, OP ⊥ AB. 
          Now AB is a chord of the circle C1 and OP ⊥ AB. 
          Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
          i.e., AP = BP
    

Example-2 :-  Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠ PTQ = 2 ∠ OPQ.

  Given that : A circle with centre O, an external point T and two tangents TP and TQ
               to the circle, where P, Q are the points of contact.
  Prove that : ∠ PTQ = 2 ∠ OPQ.
  Proof : Let ∠ PTQ = θ
          Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.
          Therefore, ∠ TPQ = ∠ TQP = 1/2 (180° - θ) = 90° - θ/2
          Also, by Theorem 10.1, ∠ OPT = 90° 
          So, ∠ OPQ = ∠ OPT – ∠ TPQ 
                    = 90° - (90° - θ/2) 
                    = θ/2 
                    = ∠ PTQ/2
          Hence, ∠ OPQ = ∠ PTQ/2
          Therefore, ∠ PTQ = 2 ∠ OPQ.
    

Example-3 :-  PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

 
  Join OT. Let it intersect PQ at the point R. Then Δ TPQ is isosceles and TO is the
  angle bisector of ∠ PTQ. So, OT ⊥ PQ and therefore, OT bisects PQ which gives
  PR = RQ = 4 cm.
  Also, OR = √OP² - PR²
           = √5² - 4²
           = 3 cm
  Now, ∠ TPR + ∠ RPO = 90° = ∠ TPR + ∠ PTR (Why?)
  So, ∠ RPO = ∠ PTR
  Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
  This gives, 
  TP/PO = RP/RO
  TP/5 = 4/3
    TP = 20/3 cm