Answer :
Step1: Draw a line PQ = 9 cm. taking P and Q as centres draw
circles of radii 5 cm and 3 cm.
Step2: Now bisect PQ. We get midpoint to be T.
Now take T as a center , draw a circle of PT radius , this will intersect
the circle at points A, B, C, D. Join PB, PD, AQ, QC.
Justification:
It can be justified by proof that PB, PD are tangents of circle (whose centre is P and radius is 5 cm) and AQ, QC are tangents of circle (whose centre is Q and radius is 3 cm)
Join PA, PC, QB, QD
∠PBQ=90° (Angle is on semicircle)
BQ ⊥ PB
Since, BQ is radius of circle, PB has to be a tangent. Similarly, PD, QA, QC are tangents.
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