Let $O$ be the centre of a circle having radius $10$ unit. Let $P$ be fixed point inside the circle such that $OP=6$. The number of chords passing through $P$ and having integral will be (Diameter is also a chord).

Using the property that a chord at a distance d from the center of a circle of radius R has length L = 2 * sqrt(R^2 - d^2), we substitute the given values to get L = 2 * sqrt(10^2 - 6^2) = 2 * sqrt(100 - 36) = 2 * sqrt(64) = 16. Since the shortest chord length through P is 16 and the longest chord is the diameter at 20, the only possible integer lengths are 16, 17, 18, and 19, giving exactly 4 such chords.

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