Two chords

Calculate the length of chord AB and perpendicular chord BC to circle if AB is 4 cm from the center of the circle and BC 8 cm from the center of the circle.

Result

a =  16 cm
b =  8 cm

Solution:

h1=4 cm h2=8 cm   r2=h12+h22  r=h12+h22=42+82=4 5 cm≐8.9443 cm  D=2⋅ r=2⋅ 8.9443=8 5 cm≐17.8885 cm  a<D,b<D  r2=h12+(a/2)2   a=2⋅ r2−h12=2⋅ 8.94432−42=16=16  cm h_{ 1 } = 4 \ cm \ \\ h_{ 2 } = 8 \ cm \ \\ \ \\ \ \\ r^2 = h_{ 1 }^2 + h_{ 2 }^2 \ \\ \ \\ r = \sqrt{ h_{ 1 }^2 + h_{ 2 }^2 } = \sqrt{ 4^2 + 8^2 } = 4 \ \sqrt{ 5 } \ cm \doteq 8.9443 \ cm \ \\ \ \\ D = 2 \cdot \ r = 2 \cdot \ 8.9443 = 8 \ \sqrt{ 5 } \ cm \doteq 17.8885 \ cm \ \\ \ \\ a<D, b<D \ \\ \ \\ r^2 = h_{ 1 }^2 + (a/2)^2 \ \\ \ \\ \ \\ a = 2 \cdot \ \sqrt{ r^2 - h_{ 1 }^2 } = 2 \cdot \ \sqrt{ 8.9443^2 - 4^2 } = 16 = 16 \ \text { cm }
r2=h22+(b/2)2  b=2⋅ r2−h22=2⋅ 8.94432−82=8=8  cm r^2 = h_{ 2 }^2 + (b/2)^2 \ \\ \ \\ b = 2 \cdot \ \sqrt{ r^2 - h_{ 2 }^2 } = 2 \cdot \ \sqrt{ 8.9443^2 - 8^2 } = 8 = 8 \ \text { cm }

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