Q104 / 180 · KCET 2010 · Mathematics
The points $(1,0),(0,1),(0,0)$ and $(2 \mathrm{k}, 3 \mathrm{k}), \mathrm{k} \neq 0$ are concyclic, if $\mathrm{k}$ is
- A$\frac{1}{5}$
- B$-\frac{1}{5}$
- C$-\frac{5}{13}$
- Correct answerD$\frac{5}{13}$
Solution
The equation of the circle which passes through the points $(1,0),(0,1)$ and $(0,0)$ is
$$
x^{2}+y^{2}-x-y=0 \quad \text{...(i)}
$$
Given that, the point $(2 \mathrm{k}, 3 \mathrm{k})$ is on the circle and form concyclic circle. Then, it satisfies the Eq. (i)
$\begin{array}{lr} & (2 \mathrm{k})^{2}+(3 \mathrm{k})^{2}-(2 \mathrm{k})-(3 \mathrm{k})=0 \\ \Rightarrow & 4 \mathrm{k}^{2}+9 \mathrm{k}^{2}-5 \mathrm{k}=0 \\ \Rightarrow & 13 \mathrm{k}^{2}-5 \mathrm{k}=0 \\ \Rightarrow & \mathrm{k}(13 \mathrm{k}-5)=0 \\ & \Rightarrow \\ \text { Hence, } & \mathrm{k}=0 \text { or } \mathrm{k}=\frac{5}{13} \\ & \mathrm{k}=\frac{5}{13}\end{array}$
Hence, the correct answer is option: 'D'.