JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 12)

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

$\frac{3}{13}$

$\frac{2}{13}$

$\frac{5}{13}$

$\frac{1}{13}$

$$(2 k, 3 k)$$ will lie on circle whose diameter is $$A B$$.

JEE Main 2024 (Online) 27th January Morning Shift Mathematics - Circle Question 24 English Explanation

$$\begin{aligned} & (x-1)(x)+(y-1)(y)=0 \\ & x^2+y^2-x-y=0 \quad \text{.... (i)} \end{aligned}$$

Satisfy (2k, 3k) in (i)

$$\begin{aligned} & (2 k)^2+(3 k)^2-2 k-3 k=0 \\ & 13 k^2-5 k=0 \\ & k=0, k=\frac{5}{13} \\ & \text { hence } k=\frac{5}{13} \end{aligned}$$