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In Fig. AB is tangent at A to the circle with centre O; point D is interior to the circle and DB intersects the circle at C. If BC = DC = 3, OD = 2 and AB = 6, then the radius of the circle is

(A)  3 + √3

(B)  √22

(C)  2√6

(D)   9/2

1 Answer

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Best answer

Correct option (B)  √22

See Fig.

BC x BE = AB2 ⇒ 3 x (6 + DE) = 36

Therefore, DE = 6.

Now DE x DC = DG x DF, so

6 x 3 = (r + 2) (r - 2) ⇒ r2 = 22

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