Let the system of equations: 2 x+3 y+5 z=9 , 7 x+3 y-2 z=8 , 12 x+3 y-(4+λ) z=16-μ . have infinitely many solutions. Then the radius of the circle centred at (λ, μ) and touching the line 4 x=3 y is

Mathematics · Determinants

JEE Main 2025 — 7 April, Morning Shift — Question 40

Let the system of equations: $2 x+3 y+5 z=9$ , $7 x+3 y-2 z=8$ , $12 x+3 y-(4+\lambda) z=16-\mu$ .

have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4 x=3 y$ is

$\Delta=\left|\begin{array}{ccc}2 & 3 & 5\\ 7 & 3 & -2\\ 12 & 3 & -(4+\lambda)\end{array}\right|$

$=2(-12-3 \lambda+6)-3(-28-7 \lambda+24)+5(21-36)$

$=-12-6 \lambda+12+21 \lambda-75$ $=15 \lambda-75$

$\Rightarrow 15 \lambda-75=0$

$\Rightarrow \lambda=5$

$\Delta_{1}=\left|\begin{array}{ccc}9 & 3 & 5\\ 8 & 3 & -2\\ 16-\mu & 3 & -9\end{array}\right|$

$=9(-27+6)-3(-72+32-2 \mu)+5(24-48+3 \mu)$

$=-189+120+6 \mu-120+15 \mu$

$=21 \mu-189=0$

$\Rightarrow \mu=9$

$r=\left|\frac{4(5)-3(9)}{\sqrt{(4)^{2}+(3)^{2}}}\right|$

$r=\frac{7}{5}$

Answer key and solution verified before publishing.

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