Perpendicular Chords in a Circle

Oct 2013
713
91
New York, USA
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.
 

Attachments

Feb 2010
738
162
Google "power of a point".
 
Oct 2013
713
91
New York, USA
I learned power of a point from videos. I used a website to graph circles and chords, and y/z doesn't have to be a constant ratio, so my proof that all three equations can't be satisfied is correct. y/z has to be a constant ratio if the chord intersects a diameter, but not if it intersects a chord. A diameter is a simpler special case of a chord. For two diameters, w, x, y, and z are equal to the radius, so 4r^2 = w^2 + x^2 + y^2 + z^2 becomes r^2 + r^2 + r^2 +r^2 = 4r^2

(x-2)^2 + (y-0.5)^2 = 16.25 is the equation of the circle. For a vertical chord on the left side of the circle with a length of 3.5, I calculated where it would intersect the circle after some failed attempts. Writing the equation of the circle in another form gives x^2 - 4x + y^2 - y = 12. To get two values of y that were 3.5 apart with the same unknown x, I made y^2 - y = (y+3.5)^2 - y + 3.5. This gave y = -1.75, which is half of -3.5. Putting it into the (y-0.5)^2 term in center and radius form gives y = -1.25 and y = 2.25. 1.75^2 = 3.0625, and 16.25 - 3.0625 = 13.1875, so (x-2)^2 = 13.1875, x - 2 = about -3.63 (I took the negative square root because I'm looking at the left side when x is negative), and x = about -1.63.

Is there a procedure for using the equation of a circle and the length of a horizontal or vertical chord to find the points where it intersects the circle? If you weren't restricting it to one side of the circle, there would be two horizontal chords and two vertical chords for each possible length. Here's a simple example. Take a circle centered at the origin with a radius of 2. Draw two horizontal chords and two vertical chords of length 1. What are the coordinates of the eight points of intersection between each chord and the circle? At first I thought that the chords would intersect on the circle to inscribe a square and have four points of intersection, but then I realized that if a square is centered at the center of the circle, there is only one possible side length of the square that has its corners on the circle, so I can't make up any chord length and have the chords intersect each other on the circle. For chords of length 1, they would not intersect on the circle. If they were lines, they would intersect outside the circle.
 
Feb 2010
738
162
In your original picture, call A the point where w,x,y,z all come together to form a right angle. Then the power of point A with respect to the circle is 12. You said that you moved the vertical chord left. As soon as you do that, the power changes. If you move the chord left, w gets smaller and x gets bigger. Yes the sum of the new w and the new x will still be 8 but their product might no longer be 12. For example, if you move the vertical chord so that w = 1 and x = 7, then the power of the point where the chords intersect is now 7 and the new y times the new z will also have a product of 7.
 
Oct 2013
713
91
New York, USA
I figured all of that out, and I figured out my question. The perpendicular bisector of a vertical chord will be a horizontal diameter and vice-versa. If I had thought of that, I would have done less work last night. I don't have anything else I'm trying to learn in this topic.
 
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