Question

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

• A. $k<0$
• B. $0<k<1$
• C. $k=1$
• D. $k>1$

Answer 1

The correct option is C $k=1$

Since, join of $(2,0)$ and $(0,3)$ subtends ${90}^{\circ }$ at $(0,0)$
$\Rightarrow$ It is a diameter.

$\therefore$ Equation is $(x-2)(x-0)+(y-0)(y-3)=0$
$x^2+y^2-2x-3y=0$

$(2k,3k)$ lies on it
$\Rightarrow 4k^2+9k^2-4k-9k=0$
$\Rightarrow 13k^2=13k$
$\Rightarrow k=1$

Since, $k\ne 0$ otherwise $(2k,3k)$ will be $(0,0)$.