Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to:

Question

Four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(1,0),(0,1)\) and \((0,0)\) lie on a circle for \(\mathrm k\) equal to :

(1) \(\frac{2}{13}\)

(2) \(\frac{3}{13}\)

(3) \(\frac{5}{13}\)

(4) \(\frac{1}{13}\)

Answer 1

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 \(\mathrm{k}=\frac{5}{13}\)