Question

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

• A. only one value of $k$
• B. $0<k<1$
• C. $k<0$
• D. all integral values of $k$

Answer 1

Answer: (A)

The equation of the circle through $\left(1,0\right)$, $\left(0,1\right)$
and $\left(0,0\right)$ is $x^2+y^2-x-y=0$
It passes through $\left(2k,3k\right)$.
So, $4k^2+9k^2-2k-3k=0$
or $13k^2-5k=0$
$\Rightarrow k\left(13k-5\right)=0$
$\Rightarrow k=0$
or $k=\frac{5}{13}\cdot$
But, $k\ne 0$ [$\because$ all the four points are distinct]
$\therefore k=\frac{5}{13}\cdot$