Question

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer 1

here we will use the method of assumes that statement is wrong. so we will suppose it is not perpendicular.take a point Q on XY other than P & join OQ. the point Q lie outside the circle that if Q lies inside the circle XY becomes secant not a tangent.
therefore OQ is longer than OP so it is true that OP is shortest of all points.so our assumption is wrong .

Answer 2

Given : A circle C(0 ,r) and a tangent L at point A

To Prove :   OA is perpendicular to L

CONSTRUCTION : Take a point B , other than A ,on the tangent L .Join OB .
                 Suppose OB meets the circle in C.
PROOF : We know that , among all line segment  joining the point O to a point on L , the perpendicular is shortest to L .

  OA = OC ( Radius of same circle)
  Now OB = OC + BC
Therefore OB greater than OC
             ⇒ OB greater than OA
             ⇒ OA is shorter than OB

 B is an arbitrary point on  the tangent L. Thus OA is shorter than any other line segment joining O to any point on L . Hence here OA is perpendicular to L.