Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and OP is joined.
Step IV: Perpendicular bisector of OP is drawn which cuts it at M.
Step V: With M as a centre and OM as a radius, a circle is drawn which intersect the circle of radius 4 cm at Q and R.
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement:
OQ = 4 cm (Radius of the smaller circle)
PO = 6 cm ( Radius of the larger circle)
∠PQO=90∘ (Angle in the semi circle)
Applying Pythagoras theorem in
ΔPQO,
PQ2+QO2=PO2 ⇒PQ2+42=62 ⇒PQ2+16=36 ⇒PQ2=36−16 ⇒PQ2=20 ⇒PQ=2√5cm Justification:
∠PQO=90∘ (Angle in the semi circle)
∴OQ⊥PQ Therefore, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.