I just learned about perpendicular chords in a circle, and I attached a screen shot. The product of the two parts of any intersection chords are equal, so y = 4. The radius is (square root of 65)/2, which is a little more than 4. A mathematician calculated that, not me. Let's say I took the vertical chord and moved it left so it remained parallel and still intersected the horizontal chord. When it is shifted, do the two parts remain in proportion? Does the part above the horizontal chord remain 4/3rds of the part below the vertical chord. If the parts remain in proportion, I can move that chord left to make y = 2 and z = 1.5. The horizontal chord would have to satisfy these three equations:
1. w + x = 8 because the length remained the same
2. wx = 2*1.5 = 3
3. w^2 + x^2 = 58.75 because the radius remained the same and 4r^2 = 65 so 4r^2  y^2  z^2 = 58.75
Satisfying the first two cannot be done with w and x > or = 1 because they would both have to be < or = 3 to make the product = 3, and therefore they can't add to 8. w and x can't both be 4, so w doesn't = x, so make w < x. Since w < 1, x must be > 7 to make the sum = 8. Since x is > 7, w < 3/7 to make the product = 3. Since w < 3/7, x > 7 4/7 to make the sum = 8. (3/7)^2 = a little more than 0.18, so let w^2 < or = 0.18. x^2 > or = 58.75  0.18 which is > or = 58.57, which means x > or = about 7.65. Since x > or = 7.65, the w < or = 0.35, and the product of about 7.65*0.35 = 2.66 is < 3. Increasing x and decreasing w while keeping a sum of 8 would make the product smaller so still < 3. I made it complicated, but I hope I provided a proof that two positive numbers cannot satisfy all three equations. Since two positive numbers cannot satisfy both equations, y/z doesn't always = 4/3. If I moved the chord left to make it equidistant from the old position and the vertical line tangent to the leftmost point of the circle, what would w, x, y, and z be? I still am guessing that y/z = 4/3 and I made a mistake.
1. w + x = 8 because the length remained the same
2. wx = 2*1.5 = 3
3. w^2 + x^2 = 58.75 because the radius remained the same and 4r^2 = 65 so 4r^2  y^2  z^2 = 58.75
Satisfying the first two cannot be done with w and x > or = 1 because they would both have to be < or = 3 to make the product = 3, and therefore they can't add to 8. w and x can't both be 4, so w doesn't = x, so make w < x. Since w < 1, x must be > 7 to make the sum = 8. Since x is > 7, w < 3/7 to make the product = 3. Since w < 3/7, x > 7 4/7 to make the sum = 8. (3/7)^2 = a little more than 0.18, so let w^2 < or = 0.18. x^2 > or = 58.75  0.18 which is > or = 58.57, which means x > or = about 7.65. Since x > or = 7.65, the w < or = 0.35, and the product of about 7.65*0.35 = 2.66 is < 3. Increasing x and decreasing w while keeping a sum of 8 would make the product smaller so still < 3. I made it complicated, but I hope I provided a proof that two positive numbers cannot satisfy all three equations. Since two positive numbers cannot satisfy both equations, y/z doesn't always = 4/3. If I moved the chord left to make it equidistant from the old position and the vertical line tangent to the leftmost point of the circle, what would w, x, y, and z be? I still am guessing that y/z = 4/3 and I made a mistake.
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