# Thread:

1. ## # of Chords with Integral Length

If the Point Z is 6 units from the center of a circle who's radius is 10.
Then, what is the number of chords with integral length, that run through Z.

2. Originally Posted by warriors837
Point P is 6 units from the center of a circle of radius 10.
Compute the number of chords with integral length that pass through P.
You can, without loss of generality, assume that the center of the circle is at (0, 0) on a coordinate system and that P is at (0, 6). Now the circle can be written as $x^2+ y^2= 100$ and any line through P as y= mx+ 6. Put that into the quadratic and solve for x to determine the points where the line crosses the circle (depending on m). Find the length of the line segment between the points (still depending on m). What values of m give an integer length?

3. Positive ones?

4. So you will arrive at:

x^2+y^2=100
x^2+(mx+6)=100
x^2+mx=94

But what does m equal?

5. The longest possible chord is the passes through both P and the center and has
length = diameter
= 20 units.

apothem d is the distance from center to a chord
The shortest possible chord is perpendicular to the longest chord and has d = 6.

central angle θ of the shortest chord = 2arccos(d/r)
= 2arccos(6/10)
= 106.26°
chord length c = 2·r·sin(θ/2)
= 2·10·0.8
= 16

You can construct a chord of any length from 16 through 20 by changing its slope.
There are five integral chord lengths from 16 through 20: 16, 17, 18, 19, 20

Correct?
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