In Fig. BDC is a tangent to the given circle at point D such that BD =

1.	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.
Answer» 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

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.

Answer»

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