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Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: A circle with centre O and radius $r=OA$ and a tangent P on point A.
To do: To prove that $OA\bot P$
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
$\therefore OA\bot P$.
Hence proved that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Updated on: 10-Oct-2022
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