Question

the tangent at any point of a circle is perpendicular to the radius through the points of contact

Answer 1

let xy is tangent to circle at point P
take point Q on xy other than P
OP=OR
OQ=OR+RQ
OP=OQ
since OP is smallest of all the distance of point O Hence OP pendicular XY

Answer 2

Answer:

$here \: is \: your \: answer$

Step-by-step explanation:

GIVEN:-

We are given a circle with centre O and a tangent XY to the circle at a point P.

TO PROVE:-

We need to prove that OP is perpendicular to XY.

PROOF:-

Take a point Q on XY other than P and join OQ as given in figure.

the point Q must lie outside the circle.

(Note : that if Q lies inside the circle, XY will become Secant and not a tangent to the circle ).

Therefore OQ is longer than the radius OP of the circle.

That Is ,

$OQ \: > \: OP$

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.