JEE Mathematics 2026 Question with Solution

Given: $|\vec{c}+\vec{d}| = \sqrt{29}$ and $\vec{c} \times (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k}) \times \vec{d}$.

Find: The value of $K$ for which $K^2x^2+(K^2-5K+\lambda_1)xy+(3K+\lambda_2^2)y^2-8x+12y+\lambda_2 = 0$ represents a circle.

Let

The given relation is

Using $\vec{v} \times \vec{d} = -(\vec{d} \times \vec{v})$,

so

Hence $\vec{c}+\vec{d}$ is parallel to $\vec{v}$, therefore

for some scalar $m$.

Now use the magnitude condition:

So the possible values are

Compute

If $m=1$, then

If $m=-1$, then

Since $\lambda_1 > \lambda_2$,

Substitute these into the conic equation:

For a second-degree equation

to represent a circle, we need

First apply the condition that the coefficient of $xy$ must be zero:

So

Now apply the condition $A=C$:

Checking the two values from the first condition:

• For $K=1$, $1-3-16=-18 \neq 0$.
• For $K=4$, $16-12-16=-12 \neq 0$.

Thus the two circle conditions are inconsistent for the printed equation, which indicates a flaw or typo in the question statement.

The solution explicitly states that the exam's intended correct option is C, and among the values obtained from the necessary condition $B=0$, that corresponds to $K=1$.

Therefore, the correct option is C, that is, $K=1$.

Given: The computed values are $\lambda_1=4$ and $\lambda_2=-4$.

Find: Whether the resulting conic satisfies all circle conditions.

After substitution, the equation becomes

Here

For a circle, both necessary conditions are:

1. $B=0$
2. $A=C$

From $B=0$,

which gives

But from $A=C$,

Neither $K=1$ nor $K=4$ satisfies this.

Therefore, the algebra extracted from the solution shows that the printed conic does not represent a circle for any listed option if both conditions are enforced exactly. The answer C is therefore the intended exam key rather than a fully consistent consequence of the printed equation.