Question

B





30cm





E D

7cm

A F C



BDC is a tangent to the given circle at point D such that BD= 30cm and CD =7cm. The other tangents BE and CF are drawn respectively from B and C to the circle and meet when produced at A makingBAC right angled triangle. Calculate (1) AF (11) radius of the circle.

Answer 1

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�+�x2�
⇒ 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