In the given figure, DE and DF are tangents from an external point D to a

It is given that DE and DF are tangents from an external point D to a circle with centre A. DE = 5 cm and DE ⊥ DF.

Join AE and AF.

Now, DE is a tangent at E and AE is the radius through the point of contact E.

$\therefore \angle \mathrm{AED}=90^{\circ} \quad$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Also, DF is a tangent at F and AF is the radius through the point of contact F.

$\therefore \angle \mathrm{AFD}=90^{\circ} \quad$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle \mathrm{EDF}=90^{\circ} \quad(\mathrm{DE} \perp \mathrm{DF})$

Also, DF = DE        (Lengths of tangents drawn from an external point to a circle are equal)

So, AEDF is a square.

∴ AE = AF = DE = 5 cm         (Sides of square are equal)

Thus, the radius of the circle is 5 cm.

Hence, the correct answer is option B.