JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 11)
Let the system of equations :
$$ \begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}$$
have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4 x=3 y$ is :
Explanation
$$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{array}\right| \\ & =2(-12-3 \lambda+6)-3(-28-7 \lambda+24)+5(21-36) \\ & =-12-6 \lambda+12+21 \lambda-75 \\ & =15 \lambda-75 \\ & \Rightarrow 15 \lambda-75=0 \\ & \Rightarrow \lambda=5 \\ & \Delta_1=\left|\begin{array}{ccc} 9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9 \end{array}\right| \\ & =9(-27+6)-3(-72+32-2 \mu)+5(24-48+3 \mu) \\ & =-189+120+6 \mu-120+15 \mu \\ & =21 \mu-189=0 \\ & \Rightarrow \mu=9 \end{aligned}$$
$$\begin{aligned} & \therefore r=\left|\frac{4(5)-3(9)}{\sqrt{(4)^2+(3)^2}}\right| \\ & r=\frac{7}{5} \end{aligned}$$
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