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प्रश्न
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Prove that the perpendicular at the point of contact to a circle passes through the centre of the circle.
उत्तर १
Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.
We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.
As perpendicular to AB at P passes through O’, therefore,
∠OPB = 90° … (1)
O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.
∴ ∠OPB = 90° … (2)
Comparing equations (1) and (2), we obtain
∠OPB = ∠OPB … (3)
From the figure, it can be observed that,
∠OPB < ∠OPB … (4)
Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.
Therefore, the perpendicular to AB through P passes through centre O.
उत्तर २
Let O be the centre of the circle.
A tangent PR has been drawn, touching the circle at point P.
Draw QP ⊥ RP at point P, such that Q lies on the circle.
∠OPR = 90° ...(Radius ⊥ tangent)
Also, ∠QPR = 90° ...(Given)
∴ ∠OPR = ∠QPR
Now, the 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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