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.

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- Oct 13th 2009, 04:25 AMwarriors837# 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. - Oct 13th 2009, 04:37 AMHallsofIvy
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 $\displaystyle x^2+ y^2= 100$ 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?

- Oct 13th 2009, 05:06 AMwarriors837
Positive ones?

- Oct 13th 2009, 10:48 AMwarriors837
So you will arrive at:

x^2+y^2=100

x^2+(mx+6)=100

x^2+mx=94

But what does m equal? - Oct 14th 2009, 07:19 PMwarriors837
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?