Question

In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is



(a) 3 cm (b) 5 cm (c) 4 cm (d) 6 cm [CBSE 2013]

Answer 1

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.

$\therefore \angle \mathrm{AED}=90°$ (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.

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

$\angle \mathrm{EDF}=90°$ (DE ⊥ 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.