# In Fig, 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 isA. 3 cmB. 5 cmC. 4 cmD. 6 cm

Given:

DE = 5 cm

DE DF

Join AE and AF

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a quadrilateral = 360°.

By property 1,

EF = ED = 5 cm(tangent from E)

And,

By property 2, AED = 90° and AFD = 90°.

Also,

EDF = 90° [ EDEF]

By property 3,

AED + AFD + EDF + EAF = 360°

90° + 90° + 90° + EAF = 360°

EAF = 360° - (90° + 90° + 90°)

EAF = 360° - 270°

EAF = 90°

All angles are equal and adjacent sides are equal AEDF is a square.

Hence, all sides are equal

AE = AF = ED = EF = 5 cm

Hence, Radius of circle = 5 cm

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