JEE Trick: Infinite Solutions → Circle Radius in 1 Minute! 🔥
❓ Question 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 \begin{cases} 2x + 3y + 5z = 9 \\ 7x + 3y - 2z = 8 \\ 12x + 3y - (4 + \lambda)z = 16 - \mu \end{cases} have infinitely many solutions . Then the radius of the circle centred at ( λ , μ ) (\lambda,\mu) and touching the line 4 x = 3 y 4x = 3y is equal to ? 🖼️ Question Image ✍️ Short Solution This question beautifully mixes: ✔ Condition for infinitely many solutions in 3 variables ✔ Linear dependence of equations ✔ Distance of a point from a line (circle tangent condition) Let’s crack it step-by-step. 🔹 Step 1 — Infinite solutions condition A system of 3 linear equations in 3 variables has infinitely many solutions 👉 iff the third equation is a linear combination of the first two (so rank < 3). So we assume ( 3 ) = a ( 1 ) + b ( 2 ) (3) = a(1) + b(2) That is: 12 x + 3 y − ( 4 + λ ) z = 16 − μ 12x + 3y - (4+\lambda)z = 16-\mu must eq...