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The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof for the theorem:
Consider a circle with centre \(O\).
Let \(AB\) be the tangent to the circle at the point \(P\).
To prove:
The line \(OP\) is perpendicular to \(AB\).
Take a point \(Q\) other than \(P\) on the tangent \(AB\) and join \(OQ\).
Here, \(Q\) must lie outside the circle.
Thus, \(OQ\) is longer than \(OP\).
That is \(OQ\) \(>\) \(OP\) at every point on \(AB\) except at \(P\).
Therefore, the point \(P\) is at the shortest distance from the centre \(O\).
We know that:
Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.
By the theorem, \(OP\) is perpendicular to \(AB\).
Hence, the proof.
In the above given figure if \(OP\) \(=\) \(3\) \(cm\)  and \(PQ\) \(=\) \(4\) \(cm\), find the length of \(OQ\).
By the result, \(\angle OPQ\) \(=\) \(90^{\circ}\).
So, \(OPQ\) is a right angled triangle.
By the Pythagoras theorem, we have:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
\(OQ^2\) \(=\) \(OP^2\) \(+\) \(PQ^2\)
\(OQ^2\) \(=\) \(3^2\) \(+\) \(4^2\)
\(OQ^2\) \(=\) \(9 + 16\)
\(OQ^2\) \(=\) \(25\)
\(\Rightarrow\)  \(OQ\) \(=\) \(\sqrt{25}\)
\(OQ\) \(=\) \(5\)
Therefore, the measure of \(OQ\) \(=\) \(5\) \(cm\)
  • By this theorem, we can say that at any point on the circle, there can be one and only one tangent.
  • The line containing the radius through the point of contact is also called the normal to the circle at the point.