Let, O be the centre of circle and AB is tangent at P. We have to prove that the perpendicular at P to AB, passes through O. Let the Perpendicular drawn at P point of AB does not pass throught O. It passes through O'. Join and O'P. Tangent drawn at P passes throught O' therefore, ∠O'PB = 90° ... (1)Read more
Let, O be the centre of circle and AB is tangent at P.
We have to prove that the perpendicular at P to AB, passes through O.
Let the Perpendicular drawn at P point of AB does not pass throught O.
It passes through O’. Join and O’P.
Tangent drawn at P passes throught O’ therefore,
∠O’PB = 90° … (1)
We know that the radius is perpendicular to tangent.
Therefore, ∠OPB = 90° … (2)
Comparing equation (1) and (2), we have
∠O’PB = ∠OPB … (3)
From figure, it is clean that,
∠O’PB < ∠OPB … (4)
Therefore, ∠O'PB = ∠OPB is not possible. It is possible only when OP and O'P coincident lines.
Therefore, the perpendicular drawn at P passes through the centre O.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the Centre.
Let, O be the centre of circle and AB is tangent at P. We have to prove that the perpendicular at P to AB, passes through O. Let the Perpendicular drawn at P point of AB does not pass throught O. It passes through O'. Join and O'P. Tangent drawn at P passes throught O' therefore, ∠O'PB = 90° ... (1)Read more
Let, O be the centre of circle and AB is tangent at P.
See lessWe have to prove that the perpendicular at P to AB, passes through O.
Let the Perpendicular drawn at P point of AB does not pass throught O.
It passes through O’. Join and O’P.
Tangent drawn at P passes throught O’ therefore,
∠O’PB = 90° … (1)
We know that the radius is perpendicular to tangent.
Therefore, ∠OPB = 90° … (2)
Comparing equation (1) and (2), we have
∠O’PB = ∠OPB … (3)
From figure, it is clean that,
∠O’PB < ∠OPB … (4)
Therefore, ∠O'PB = ∠OPB is not possible. It is possible only when OP and O'P coincident lines.
Therefore, the perpendicular drawn at P passes through the centre O.