Correct option is (3) \(\frac{5}{13}\)
(2k, 3k) will lie on circle whose diameter is \(\mathrm{AB}\).
\((\mathrm{x}-1)(\mathrm{x})+(\mathrm{y}-1)(\mathrm{y})=0\)
\(\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}=0 \quad ....(i)\)
Satisfy (2k, 3k) in (i)
\(\mathrm{(2 k)^{2}+(3 k)^{2}-2 k-3 k=0}\)
\(13 \mathrm{k}^{2}-5 \mathrm{k}=0\)
\(\mathrm{k}=0, \mathrm{k}=\frac{5}{13}\)
Hence \(\mathrm{k}=\frac{5}{13}\)