Theory:

Result \(1\):
A tangent at any point on a circle and the radius through the point are perpendicular to each other.
Explanation:
 
r1.png
 
The tangent at the point \(P\) on a circle and the radius through the point \(P\) are perpendicular to each other.
 
That is, the radius \(OP\) makes an angle \(90^{\circ}\) with the tangent \(AB\) at the point \(P\).
Example:
In the above given figure if \(OP\) \(=\) \(3\) \(cm\)  and \(PQ\) \(=\) \(4\) \(cm\), find the length of \(OQ\).
 
Solution:
 
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 \(OQ\) \(=\) \(5\) \(cm\)
Result \(2\):
  • No tangent can be drawn from an interior point of the circle.
r2_1.png
  • Only one tangent can be drawn at any point on a circle.
r2_2.png
  • Two tangents can be drawn from any exterior point of a circle.
r2_3.png