Working rule to construct a pair of tangents to a circle from an external point \(P\):
Given the radius of a circle and the distance of the external point from the centre, let us learn how to construct a pair of tangents to the circle.
Draw a circle of diameter \(6\) \(cm\) from a point \(P\), which is \(8\) \(cm\) away from its centre. Draw the two tangents \(PA\) and \(PB\) to the circle and measure their lengths.
The diameter of the circle \(=\) \(6\) \(cm\).
Radius of the circles \(=\) \(\frac{Diameter}{2}\)
Radius \(=\) \(\frac{6}{2}\)
\(=\) \(3\) \(cm\)
Rough Sketch:
Step \(1\): With \(O\) as the centre, draw a circle of radius \(3\) \(cm\).
Step \(2\): Draw a line \(OP\) of length \(8\) \(cm\).
Step \(3\): Draw a perpendicular bisector of \(OP\), which cuts \(OP\) at \(M\).
Step \(4\): With \(M\) as centre and \(MO\) as radius, draw a circle that cuts the previous circle at \(A\) and \(B\).
Step \(5\): Join \(AP\) and \(BP\). \(AP\) and \(BP\) are the required tangents. Thus length of the
tangents are \(PA = PB = 7.4\) \(cm\).
In the right angle triangle \(OAP\) by the Pythagoras theorem, we have:
\(OP^2\) \(=\) \(OA^2\) \(+\) \(PA^2\)
\(\Rightarrow\) \(PA^2\) \(=\) \(OP^2\) \(-\) \(OA^2\)
\(PA^2\) \(=\) \(8^2\) \(+\) \(3^2\)
\(PA^2\) \(=\) \(64 – 9\)
\(PA^2\) \(=\) \(55\)
\(\Rightarrow PA = \sqrt{55}\)
\(PA\) \(=\) \(7.4\) \(cm\) (approximately) .