Theorem-1 :-  The tangent at any point of a circle is perpendicular to the radius through the point of contact.

        
  Given that : A circle with centre O and a tangent XY to the circle at a point P. 
  Prove that : OP is perpendicular to XY.
  Construction : Take a point Q on XY other than P and join OQ.
  Proof : The point Q must lie outside the circle. 
          Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). 
          Therefore, OQ is longer than the radius OP of the circle. i.e., OQ > OP. 
          Since this happens for every point on the line XY except the point P, 
          OP is the shortest of all the distances of the point O to the points of XY. 
          So OP is perpendicular to XY. 
    

Theorem-2 :-  The lengths of tangents drawn from an external point to a circle are equal.

        
  Given that : A circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P.  
  Prove that : PQ = PR.
  Construction : we join OP, OQ and OR.
  Proof : ∠ OQP and ∠ ORP are right angles, because these are angles between the radii 
          and tangents, and according to Theorem 10.1 they are right angles. 
          Now in right triangles OQP and ORP, 
          OQ = OR (Radii of the same circle) 
          OP = OP (Common) 
          Therefore, Δ OQP ≅ Δ ORP (RHS) 
          This gives PQ = PR (CPCT).