Bunuel wrote:

In the figure above, O is the center of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has radius of 2, then AC =

(A) 4

(B) 2√2

(C) 4 + √2

(D) 4 + √3

(E) 2 + 2√2

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The length of AC = (length of hypotenuse, AO, of an isosceles right triangle) + (radius, CO)

Find length of AOConnect O with B

Point B is tangent, so the radius is perpendicular to line AB and creates a right angle at point B

The triangle hence is a 45-45-90 isosceles right triangle, with side lengths in ratio*

\(x: x: x\sqrt{2}\)One leg, BO \(= x\) =

radius =

\(2\) Other leg, AB = BO =

\(x = r = 2\)Hypotenuse, AO = \(x\sqrt{2}=\)

\(2\sqrt{2}\)Add segment CO to get length of ACThe rest of the length of AC is

CO =

r =

\(2\)Length of AC =

\(2\sqrt{2} + 2\)Answer E

*

OR Pythagorean theorem:

\(leg^2 + leg^2 = hypotenuse, h^2\)

\(2^2 + 2^2 = h^2\)

\(8 = h^2\)

\(\sqrt{4*2} = \sqrt{h^2}\)

\(h = 2\sqrt{2}\) = length of AO
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