Answer is D. 5 cm
Given:
SQ = 6 cm
QR = 4 cm
Property: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.
By above property, ∆ROQ is right-angled at ∠OQR (i.e., ∠OQR = 90°).
Diameter QS = 6 cm
Radius = \(\frac{Diameter }{2}\)
⇒Radius = \(\frac{6cm}{2}\)
⇒Radius (OQ) = 3 cm
Now by Pythagoras theorem,
OR2 = OQ2 + QR2
⇒ OR2 = 32 + 42
⇒ OR2 = 9+ 16
⇒ OR2 = 25
⇒ OR= √25
⇒ OR= 5 cm
Hence, OR= 5 cm.