Confirm that the tangent to a circle is at right angles to the radius through the point of contact.
Given:
Let \(O\) be the centre of the circle.
\(PT\) is a tangent to the circle at point \(P\).
To prove: \(OP\perp PT\).
Construction: Take an point \(Q\) on the tangent \(PT\) other than \(P\) and join \(OQ\).
Proof:
Since \(O\) is the centre of the circle and \(P\) lies on the circle,
\(= r\)
Where \(r\) is the radius of the circle.
The point \(Q\) lies on the tangent \(PT\) but not on the circle.
Therefore, \(Q\) is an of the circle.
Hence, \(OQ\)\(OP\).
Because the distance from a point to a line segment is the distance, and every point on the tangent except \(P\) lies the circle.
Thus, among all the line segments joining \(O\) to points on the tangent \(PT\).
\(OP\)\(OQ\)
for every point \(Q\neq P\) on the tangent.
So, \(OP\) is the distance from the centre \(O\) to the tangent \(PT\).
Hence, \(OP\perp PT\).