Draw a circle of radius 3 cm. Take two points at a distance 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Steps of Construction :
(i) Bisect PQ. Let M be the mid-point of PO.
(ii)    Taking M as centre and MO as radius, draw a circle which intersect the given circle at the points A and B.
(iii)    Join PA and PB.
Now, PA and PB are the required two tangents.
(iv)    Bisect QO. Let N be the mid-point of QO.
(v)    Taking N as centre and NO as radius, draw a circle. Let it intersect the given circle at the points C and D.
(vi)    Join QC and QD.

Then QC and QD are the required two tangents.
Justification : Join OA and OB.
Then ∠PAO is an angle in the semicircle and, therefore,
∠PAO = 90°
⇒    PA ⊥ OA
Since, OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.
Again, Join OC and OD.
Then ∠QCO is an angle on the semicircle and therefore,
∠QCO = 90°
Since, OC is a radius of the given circle, QC has to be a tangent to the circle.
Similarly, QD is also a tangent to the circle.