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


Given: A tangent to a circle.

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

Solution:

Take a point B, other than point A on the tangent P.

 Join OB.

Suppose OB meets the circle on the point C.

Proof: We know that, among all line segment joining the point 0 to a point on P,

The perpendicular is shortest to P

$OA = OC\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( Radius\ of\ the\ same\ circle)$

Now, $OB\ =\ OC\ +\ BC$

$\Rightarrow \ OB\  >\ OC$

$\Rightarrow OB\  >\ OA$

$\Rightarrow OA\

B is an arbitrary point on the tangent P.

Thus, OA is shorter than any other line segment joining O to any point on P Here,

$OA\ \perp P$

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

79 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements