Let AB is diameter, PQ and RS are tangents drawn at ends of diameter. We Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ. ∠OAR = 90° and OAS = 90° ∠OBP = 90° and ∠OBQ = 90° From the above, we have ∠OAR = ∠OBQ [Alternate angles] ∠OAS = ∠OBP [Alternate angles] Since, alternate anglRead more
Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
We Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ.
∠OAR = 90° and OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal. Hence, PQ is parallel to PS.
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 tangents drawn at the ends of a diameter of a circle are parallel.
Let AB is diameter, PQ and RS are tangents drawn at ends of diameter. We Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ. ∠OAR = 90° and OAS = 90° ∠OBP = 90° and ∠OBQ = 90° From the above, we have ∠OAR = ∠OBQ [Alternate angles] ∠OAS = ∠OBP [Alternate angles] Since, alternate anglRead more
Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
See lessWe Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ.
∠OAR = 90° and OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal. Hence, PQ is parallel to PS.
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.