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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Determinants

Let the system of equations : aligned & 2 x+3 y+5 z=9 & 7 x+3 y-2...

Let the system of equations :
<br/>2x+3y+5z=97x+3y−2z=812x+3y−(4+λ)z=16−μ<br/>\begin{aligned}<br/>& 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu<br/>\end{aligned}
have infinitely many solutions. Then the radius of the circle centred at (λ,μ)(\lambda, \mu) and touching the line 4x=3y4 x=3 y is

Options

  1. A175\frac{17}{5}
  2. B75\frac{7}{5}
  3. C7
  4. D215\frac{21}{5}

Correct answer

B. 75\frac{7}{5}

Step-by-step solution

∣23573−2123−(λ+4)∣=0⇒12(−21)−3(−39)−(λ+4)(−15)=0⇒−252+117+15(1+4)=0⇒15λ+177−252=0⇒15λ−75=0⇒λ=5∣93583−216−μ3−9∣=0⇒∣107μ−80716−μ3−9∣=0⇒7−7(μ−8)=0⇒1−(μ−8)=0⇒μ=9⇒ centre of circle (5,9) radius = length of ⊥ from centre (5,9)=∣20−275∣=75\begin{aligned} & \left|\begin{array}{ccc}2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda+4)\end{array}\right|=0 \\ & \Rightarrow 12(-21)-3(-39)-(\lambda+4)(-15)=0 \\ & \Rightarrow-252+117+15(1+4)=0 \\ & \Rightarrow 15 \lambda+177-252=0 \\ & \Rightarrow 15 \lambda-75=0 \Rightarrow \lambda=5 \\ & \left|\begin{array}{ccc}9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}1 & 0 & 7 \\ \mu-8 & 0 & 7 \\ 16-\mu & 3 & -9\end{array}\right|=0 \\ & \Rightarrow 7-7(\mu-8)=0 \Rightarrow 1-(\mu-8)=0 \Rightarrow \mu=9 \\ & \Rightarrow \text{ centre of circle }(5,9) \\ & \text{ radius }=\text{ length of } \perp \text{ from centre }(5, \quad 9)= \\ & \left|\frac{20-27}{5}\right|=\frac{7}{5}\end{aligned}
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