Question

The points $(1, 0), (0, 1), (0, 0)$ and $(2k, 3k),k \neq 0$ are concyclic if $k$ = _____

• A. $-\frac {5}{13}$
• B. $\frac {5}{13}$
• C. $\frac {1}{5}$
• D. $-\frac {1}{5}$

Answer 1

Answer: (B)

$\frac {5}{13}$

Answer 2

The equation of the circle which passes through the points $(1,0),(0,1)$ and $(0,0)$ is
$x^{2}+y^{2}-x-y=0$
Given that, the point $(2 k, 3 k)$ is on the circle and form concyclic circle. Then, it satisfies the E (i)
$(2 k)^{2}+(3 k)^{2}-(2 k)-(3 k)=0$
$\Rightarrow 4 k^{2}+9 k^{2}-5 k=0$
$\Rightarrow 13 k^{2}-5 k=0$
$\Rightarrow k(13 k-5)=0$
$\Rightarrow k=0$ or $k=\frac{5}{13}$
Hence, $k=\frac{5}{13}$