Number of chords of a circle having natural length

Question

In a circle of radius 17, point p lies on a distance 15 from center.How many distinct chords of this circle passing through p do have a natural length?
I tried to use the notion of Power of point relative to the circle and found the following chord segments:
$(2,32) (4,16) (8,8) (16,4) (32,2)$ , (these chords have lengths of $ 34,20$ etc.)
But it seems there must be more such chords! How??!!

Answer 1

There is a longest chord of length $2\times 17=34$ along the diameter through the point and a perpendicular shortest chord through the point of length $2 \times \sqrt{17^2-15^2}=16$ and two chords of every intermediate length.

So the number of chords of integer length is $1+1+2\times (34-16-1)=36$

Answer 2

The longest possible chord through $p$ is the diameter through it; the shortest possible chord is the one perpendicular to the diameter. If one or both of these are of integer length (the diameter surely is, having length $34$), they count for one each.

Every integer between those two limits is the length of exactly $2$ different chords.

(The mistake in your attempt seems to be that you're restricting yourself to chords where not only the length of the chord is an integer, but where $p$ divides it in two integral pieces -- which the problem doesn't appear to require. Also, your $2+32$ and $32+2$ solutions both denote the same chord, namely the diameter through $p$ seen from either one or the other end).

Answer 3

The other two answers do not explain why every chord length between the longest and shortest chords is possible. The reason is that as you vary a point on the circle, the chord through that point and $p$ varies continuously and hence the other intersection with the circle varies continuously, and so their distance varies continuously. Then the intermediate value theorem shows that every intermediate chord length is attained.