Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.
Steps of Construction:
1. A circle with radius $$6\ cm$$ is drawn taking $$O$$ as centre
2. Point $$P$$ is marked at $$10\ cm$$ away from centre of circle.
3. With the half of compass mark $$M$$ which is the midpoint of $$OP$$.
4. Draw a circle with centre $$M$$, taking radius $$MO$$ or $$MP$$ which intersects the given circle at $$Q$$ and $$R$$.
5. Now join $$PQ$$ and $$PR$$. These are the tangents of the circle.
We know that, the tangent to a circle is perpendicular to the radius through the point of contact.
$$\therefore \ $$In $$\triangle OPQ, \ OQ\perp QP$$
and in $$\triangle OPR, \ OR\perp PR$$
Hence, both $$\triangle OPQ$$ and $$\triangle OPR$$ are right angle triangles.
Applying Pythagoras theorem to both $$\triangle s$$, we get:
$$OP^2=OQ^2+PQ^2$$
and $$OP^2=OR^2+PR^2$$
$$OQ=OR=radius=6\ cm$$ and $$OP=10\ cm$$
$$\therefore \ 10^2=6^2+PQ^2$$
and $$10^2=6^2+PR^2$$
$$\Rightarrow PQ^2=100-36=64$$
and $$PR^2=100-36=64$$
$$\therefore \ PQ=PR=8\ cm$$
Hence, the length of the tangents to a circle of radius $$6\ cm$$, from a point $$10\ cm$$ away from the centre of the circle, is $$8\ cm$$.