JEE Mathematics 2025 Question with Solution

Given: The system

$$\begin{aligned} 2x + 3y + 5z &= 9 \\ 7x + 3y - 2z &= 8 \\ 12x + 3y - (4 + \lambda)z &= 16 - \mu \end{aligned}$$

has infinitely many solutions.

Find: The radius of the circle centered at $(\lambda, \mu)$ and touching the line $4x = 3y$.

For infinitely many solutions, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4 + \lambda) \end{vmatrix} = 0$$

Expanding the determinant gives $\lambda = 5$.

Now using the determinant formed with the constants in place of the third column:

$$\begin{vmatrix} 2 & 3 & 9 \\ 7 & 3 & 8 \\ 12 & 3 & 16 - \mu \end{vmatrix} = 0$$

we get $\mu = 9$.

So the center of the circle is $(5, 9)$.

The given line is

$$4x - 3y = 0$$

The radius is the perpendicular distance from $(5, 9)$ to this line:

$$r = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + (-3)^2}}$$ $$r = \frac{|20 - 27|}{\sqrt{16 + 9}} = \frac{7}{5}$$

Therefore, the radius of the circle is $\frac{7}{5}$, so the correct option is B.

Given: The system has infinitely many solutions.

Find: The radius of the circle centered at $(\lambda, \mu)$.

Use the condition directly from the solution:

$$\begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda + 4) \end{vmatrix} = 0 \Rightarrow \lambda = 5$$

and

$$\begin{vmatrix} 2 & 3 & 9 \\ 7 & 3 & 8 \\ 12 & 3 & 16 - \mu \end{vmatrix} = 0 \Rightarrow \mu = 9$$

Hence the center is $(5, 9)$.

Now convert the line to standard form $4x - 3y = 0$ and apply the point-to-line distance formula:

$$r = \frac{|4 \cdot 5 - 3 \cdot 9|}{\sqrt{4^2 + 3^2}} = \frac{7}{5}$$

Therefore, the radius is $\frac{7}{5}$ and the correct option is B.