If four distinct points $(2 k, 3 k),(2,0),(0,3)$ (0,0) lie on a circle, then

• A

  $k < 0$
• B

  $0 < k < 1$
• Correct answer
  C

  $k=1$
• D

  $k>1$

Solution

Since, join of (2,0) and (0,3) subtends $90^{\circ}$ at (0,0) $\Rightarrow$ It is a diameter. $\therefore$ Equation is $(x-2)(x-0)+(y-0)(y-3)=0$ $x^{2}+y^{2}-2 x-3 y=0$ $(2 k, 3 k)$ lies on it $\Rightarrow \quad 4 k^{2}+9 k^{2}-4 k-9 k=0$ $\Rightarrow$ . $13 k^{2}=13 k$ $\Rightarrow$ $k=1$ since, $k \neq 0$ otherwise $(2 k, 3 k)$ will be (0,0)

Hence, the correct answer is option: 'C'.