Give also the justification of the construction:
Draw a circle of radius $6 \, cm$. From a point $10 \, cm$ away from its centre,construct the pair of tangents to the circle and measure their lengths.

(N/A) pair of tangents to the given circle can be constructed as follows:
$1.$ Taking any point $O$ of the given plane as centre,draw a circle of $6 \, cm$ radius. Locate a point $P$,$10 \, cm$ away from $O$.
Join $OP$.
$2.$ Bisect $OP$. Let $M$ be the mid-point of $PO$.
$3.$ Taking $M$ as centre and $MO$ as radius,draw a circle.
$4.$ Let this circle intersect the previous circle at points $Q$ and $R$.
$5.$ Join $PQ$ and $PR$. $PQ$ and $PR$ are the required tangents.
The lengths of tangents $PQ$ and $PR$ are $8 \, cm$ each.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $6 \, cm$). For this,join $OQ$ and $OR$.
$\angle PQO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PQO = 90^{\circ}$
$\Rightarrow OQ \perp PQ$
Since $OQ$ is the radius of the circle,$PQ$ has to be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.