JEE Mathematics 2024 Question with Solution

Given: The points $$ (2k,3k), (1,0), (0,1), (0,0)

Find: The value of $k$.

The points $(1,0), (0,1), (0,0)$ determine a unique circle. Using the circle with diameter joining $(1,0)$ and $$ (0,1)

(x-1)x + (y-1)y = 0

x^2 + y^2 - x - y = 0

Now substitute the point $$ (2k,3k)

(2k)^2 + (3k)^2 - 2k - 3k = 0

4k^2 + 9k^2 - 5k = 0

13k^2 - 5k = 0

k(13k-5) = 0

So the possible values are

Since $k=0$ makes the point $(2k,3k)=(0,0)$, it is not a distinct fourth point. Hence the valid value is $$ \frac{5}{13}

Therefore, the correct option is C.

Given: The points $$ (2k,3k), (1,0), (0,1), (0,0)

Find: The value of $k$.

A standard concyclicity test is

The first approach in the source shows this method but contains an arithmetic inconsistency in the working.

The reliable simplification comes from first finding the circle through $(1,0), (0,1), (0,0)$ and then checking whether $$ (2k,3k)

x^2 + y^2 - x - y = 0

13k^2 - 5k = 0

k = 0 nd nd k = \frac{5}{13}

Discarding $k=0$ because it repeats the point $(0,0)$, we get the required value $k=\frac{5}{13}$. The source solution's final conclusion also states $$ \frac{5}{13}

Therefore, the correct option is C.