In Fig. 10.99, 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
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,
AE = AF [radius]
By property 2, ∠AED = 90° and ∠AFD = 90°.
Also,
∠EDF = 90° [∵ ED⊥EF]
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