Question

lf $(2k,3k),(2,0),(0,3),(0,0)$ lie on a circle (the points being distinct), then

• A. $0<k<1$
• B. $k=1$
• C. $k<0$
• D. k has two values

Answer 1

The correct option is B $k=1$

Since the circle passes through $A(2,0)$ and $B(0,3)$ and
$\angle AOB=\frac{\pi }{2}$, the circle is the circle on $AB$ as diameter.
$\therefore$ its equation is $(x-2)x+(y-3)=0$
This passes though (2k,3k)
$\Rightarrow 13k^2-13k=0$
$\Rightarrow k=0$ or $1$
But the given four point are distinct
$\Rightarrow k=1.$