Question

# Two perpendicular chords AB and CD of a circle intersect at O, (inside the circle AB is bisected at O). If AO = 4 units and OD = 6 units. Find circumradius of $$\displaystyle \Delta ACO$$.

A
13
B
2133
C
152
D
2154

Solution

## The correct option is B $$\displaystyle \frac{2\sqrt{13}}{3}$$When 2 chords intersect, the product of segment of one chord is equal to the product of segments of the other $$\implies AO \times BO = OC \times OD$$ Given $$AO = BO = 4$$ $$\implies 4 \times 4 = OC \times 6$$ $$OC = \dfrac{8}{3}$$ Given $$AB$$ is perpendicular to $$CD$$ $$\implies \angle COA = 90$$ $$AC^2 = OC^2 + OA^2 = \dfrac{64}{9} + 16$$ $$AC = \dfrac{\sqrt{208}}{3}$$ Since $$AOC$$ is a right triangle $$Circumradius = \dfrac{1}{2}(hypotenuse)$$ $$= \dfrac{1}{2} \times AC = \dfrac{\sqrt{208}}{6} = \dfrac{4\sqrt{13}}{6}$$ $$r = \dfrac{2\sqrt{13}}{3}$$Maths

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