Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its center. Draw tangents to the circle from these two points P and Q.
 

Steps of construction:

1. Draw a circle with O as centre and radius as 3 cm.
2. Draw a diameter of it and extend both the sides and mark the points as P, Q such that OP = OQ = 7 cm.
3. Draw the perpendicular bisectors of OP and OQ to intersect PQ at M and N respectively.
4. With M as centre and OM as radius draw a circle to cut the given circle at A and C. With N as the centre and ON as radius draw a circle to cut the given circle at B and D.
5. Join PA, PC, QB, QD. 

PA, PC and QB, QD are the required tangents from P and Q respectively.

Proof:

∠PAO = ∠QBO = 90° (Angle in a semi-circle)

∴ PA ⊥ AO, QB ⊥ BO

In right angle triangle PAO,

OP = OQ = 7cm (By construction)

OA = OB = 3cm (radius of the given circle)

PA² = (OP)² - (OA)²

= (7)² - (3)²

= 49 - 9

= 40

PA = √40

= 6.3 (approx.)

Also, in right triangle QBO,

QB² = (OQ)² - (OB)²

= (7)² - (3)²

= 49 - 9

= 40

QB = √40

= 6.3 (approx)