Prove that a tangent to a circle is perpendicular to the radius to the

1.	Prove that a tangent to a circle is perpendicular to the radius to the point of contact. i.e in the figure given below, if AB is a tangent to the circle C(O,r) at P which is the point of contact, then OP perpendicular AB.
Answer» GIVEN: A circle C(O,r ) and a tangent XY at a point P. TO PROVE: OP ⊥ XY CONSTRUCTION: Take a point x, other than P, on the Tangent XY. Join OQ. Suppose OQ meets the circle at R. PROOF: We know that among all line segments joining the points O to a point on XY, the shortest ONE is Perpendicular to XY . So, to prove that OP ⊥ XY , it is sufficient to prove that OP is shorter than any other SEGMENT joining O to any point of XY . Clearly, ⠀⠀⠀⠀ $\sf{ OP = OR }$ [Radii of the circle] Now, ⠀⠀⠀⠀ $\sf{ OQ = OR + RQ}$ $\sf{\implies OQ > OR}$ $\sf{\implies OQ > OP}$ [ ∵ OP = OR] $\sf{\implies OP < OQ}$ THUS, OP is shorter than any other segment joining O to any point of XY. Hence, ⠀⠀⠀⠀OP ⊥ XY *⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Proved...⠀⠀⠀⠀⠀*

Prove that a tangent to a circle is perpendicular to the radius to the point of contact. i.e in the figure given below, if AB is a tangent to the circle C(O,r) at P which is the point of contact, then OP perpendicular AB.

Answer»

GIVEN:

A circle C(O,r ) and a tangent XY at a point P.

TO PROVE:

OP ⊥ XY

CONSTRUCTION:

Take a point x, other than P, on the Tangent XY. Join OQ. Suppose OQ meets the circle at R.

PROOF:

We know that among all line segments joining the points O to a point on XY, the shortest ONE is Perpendicular to XY . So, to prove that OP ⊥ XY , it is sufficient to prove that OP is shorter than any other SEGMENT joining O to any point of XY .