Let the system of equations : 2x + 3y + 5z = 9, 7x + 3y - 2z = 8, 12x + 3y – (4 + λ)z = 16 – μ, have infinitely many solutions.

Question

Let the system of equations :

2x + 3y + 5z = 9,

7x + 3y - 2z = 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 4x = 3y is

(1) \(\frac{17}{5}\)

(2) \(\frac{7}{5}\)

(3) 7

(4) \(\frac{21}{5}\)

Answer 1

Correct option is: (2) \(\frac{7}{5}\) 

\( \mathrm{D}=0 \Rightarrow\left|\begin{array}{ccc} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda+4) \end{array}\right|=0 \)

\( \lambda=5 \)

\( \mathrm{D}_3=0 \Rightarrow\left|\begin{array}{ccc} 2 & 3 & 9 \\ 7 & 3 & 8 \\ 12 & 3 & 16-\mu \end{array}\right|=0 \Rightarrow \mu=9\)

Centre (5, 9)

\(\text { radium }=\frac{|5 \times 4-3 \times 9|}{5}=\frac{7}{5}\)