### 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:

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 $$PA = PB = 7.4$$ $$cm$$.

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) .