Question

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

Answer 1

Referring to the figure:

OA=OC (Radii of circle)

Now OB=OC+BC

∴OB>OC (OC being radius and B any point on tangent)

⇒OA<OB

B is an arbitrary point on the tangent.

Thus, OA is shorter than any other line segment joining O to any

point on tangent.

Shortest distance of a point from a given line is the perpendicular distance from that line.

Hence, the tangent at any point of circle is perpendicular to the radius.

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Answer 2

To Prove:

The tangents at any point of a circle are perpendicular to the radius through the point of contact.

Solution:

Let the center of the circle be O.

The triangle formed by it AOC,

OA and OC are the radii of the circle.

So, OA = OC [Radii of the circle]

AB is the tangent of the circle.

OB = OC + BC

OB > OC

OA < OB

B is an arbitrary point on the tangent.

So we can see that OA is shorter than other lines meeting O to any point on the tangent and the shortest distance of a point from a given line is the perpendicular distance from that line.

Hence proved that the tangent at any point of the circle is perpendicular to the radius through the point of contact.