### Theory:

*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.

Example:

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.

**Given**:

The diameter of the circle \(=\) \(6\) \(cm\).

Radius of the circles \(=\) \(\frac{Diameter}{2}\)

Radius \(=\) \(\frac{6}{2}\)

\(=\) \(3\) \(cm\)

**:**

*Rough Sketch*Construction:

**: With \(O\) as the centre, draw a circle of radius \(3\) \(cm\).**

*Step \(1\)***: Draw a line \(OP\) of length \(8\) \(cm\).**

*Step \(2\)***: Draw a perpendicular bisector of \(OP\), which cuts \(OP\) at \(M\).**

*Step \(3\)***: With \(M\) as centre and \(MO\) as radius, draw a circle that cuts the previous circle at \(A\) and \(B\).**

*Step \(4\)***: Join \(AP\) and \(BP\). \(AP\) and \(BP\) are the required tangents. Thus length of the**

*Step \(5\)*tangents are \(PA = PB = 7.4\) \(cm\).

Verification:

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) .