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In Fig. BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.

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Given : AB, BC and AC are tangents to the circle at E, D and F. 

BD = 30 cm and DC = 7 cm and ∠BAC = 90° 

Recall that tangents drawn from an exterior point to a circle are equal in length 

Hence BE = BD = 30 cm 

Also FC = DC = 7 cm 

Let AE = AF = x → (1) 

Then AB = BE + AE = (30 + x) 

AC = AF + FC = (7 + x) 

BC = BD + DC = 30 + 7 = 37 cm 

Consider right Δ ABC, by Pythagoras theorem we have 

BC2 = AB2 + AC2 

⇒ (37)2 = (30 + x)2 + (7 + x)2 

⇒ 1369 = 900 + 60x + x2 + 49 + 14x + x

⇒ 2x2 + 74x + 949 – 1369 = 0 

⇒ 2x2+ 74x – 420 = 0 

⇒ x2 + 37x – 210 = 0 

⇒ x2 + 42x – 5x – 210 = 0 

⇒ x (x + 42) – 5 (x + 42) = 0 

⇒ (x – 5) (x + 42) = 0 

⇒ (x – 5) = 0 or (x + 42) = 0 

⇒ x = 5 or x = – 42 

⇒ x = 5 [Since x cannot be negative] 

∴ AF = 5 cm [From (1)] 

Therefore AB = 30 +x = 30 + 5 = 35 cm 

AC = 7 + x = 7 + 5 = 12 cm 

Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle. 

Join point O, F; points O, D and points O, E. 

From the figure, 

Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC) 

∴ r = 5 

Thus the radius of the circle is 5 cm

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