Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point.
Step IV: With O as a center and OX as its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as center and OY as radius and it intersects the circle at R and T.
Step VI: PM and PN are joined and also QR and QT are joined.
Thus, PM and PN are tangents to the circle from P. QR and QT are tangents to the circle from point Q.
Justification:
∠PMO=90∘ (Angle in the semi-circle)
∴OM⊥PM
Therefore, OM is the radius of the circle then PM has to be a tangent of the circle.
Similarly, PN, QR, and QT are tangents of the circle.