Two perpendicular chords intersect in a circle. The segments of one ch

Look at the attached image for a explanation of the variables and dimensions.

Given,

GF = 3, GE = 4, BG = 6 and CG = 2

O is the center of the circle. Chords BC and EF are perpendicular to each other.

Now, in triangle OBC, OB = OC = radius of the circle, thus triangle OBC is an isosceles triangle and OA \(\perp\) BC (OA is perpendicular to BC). Thus OA will bisect BC into 2 equal parts (a property of an isosceles triangle that the perpendicular drawn from one of the vertex to the base (assuming base is NOT one of the 2 equal sides), bisects the base).

---> AB = AC = (6+2)/2 = 4 ....(1)

Similarly in triangle OFE, OF = OE = radius of the circle. OD \(\perp\) EF ---> FD = DE = (3+4)/2 = 3.5 ---> FD = DG + FG ---> DG = FD - FG = 3.5 - 3 = 0.5 ...(2)

Also, OAGD form a rectangle such that GD = OA = 0.5 and AG = OD = 2.

Thus, in right triangle OAB,\(OA^2+AB^2 = OB^2\) --->\(r^2 = 0.5^2+4^2 = 16.25\) ---> \(r = \sqrt{16.25}\) ---> diameter = \(2r = 2 \sqrt{16.25}\) = \(\sqrt{4*16.25} = \sqrt{65}\) .

C is the correct answer.