(N/A) The tangents can be constructed on the given circle as follows:
$1.$ Taking any point $O$ on the given plane as the centre,draw a circle of $3 \, cm$ radius.
$2.$ Take one of its diameters and extend it on both sides. Locate two points $P$ and $Q$ on this diameter such that $OP = OQ = 7 \, cm$.
$3.$ Bisect $OP$ and $OQ$. Let $T$ and $U$ be the mid-points of $OP$ and $OQ$ respectively.
$4.$ Taking $T$ and $U$ as centres and $TO$ and $UO$ as radii,draw two circles. These two circles will intersect the original circle at points $V, W$ and $X, Y$ respectively. Join $PV, PW, QX,$ and $QY$. These are the required tangents.
Justification:
The construction can be justified by proving that $PV, PW, QX,$ and $QY$ are the tangents to the circle (whose centre is $O$ and radius is $3 \, cm$). For this,join $OV, OW, OX,$ and $OY$.
$\angle PVO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PVO = 90^{\circ}$
$\Rightarrow OV \perp PV$
Since $OV$ is the radius of the circle,$PV$ must be a tangent to the circle.
Similarly,it can be shown that $PW, QX,$ and $QY$ are the tangents to the circle.