prove that the perpendicular at any point of contact to the tangent to a circle passes through the center
Answers
Step-by-step explanation:
O is the centre of the given circle.
A tangent PR has been drawn touching the circle at point P.
Draw QP ⊥ RP at point P, such that point Q lies on the circle.
∠OPR = 90° (radius ⊥ tangent)
Also, ∠QPR = 90° (Given)
∴ ∠OPR = ∠QPR
Now, above case is possible only when centre O lies on the line QP.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
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Given :Let us assume a circle with center O and AB be the tangent intersecting circle at point P
To prove : OP ⊥AB
proof :
we know that tangent of a circle is perpendicular to radius at point of contact
OP ⊥AB
So ∠OPB = 90°
noe let us assume a point X
such that XP⊥ AB
∠XPB = 90°..(2)
From 1 and 2
∠OPB =∠XPB = 90°
which is possible if the line XP passes through O
hence , perpendicular to the tangent to a circle passes through the center
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https://brainly.in/question/12480022