Let the system of equations : … have infinitely many solutions. Then the radius of the circle centred at (λ, μ) and touching the line 4 x=3 y is

$$\begin{aligned} & \left|\begin{array}{ccc}2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda+4)\end{array}\right|=0 \\ & \Rightarrow 12(-21)-3(-39)-(\lambda+4)(-15)=0 \\ & \Rightarrow-252+117+15(1+4)=0 \\ & \Rightarrow 15 \lambda+177-252=0 \\ & \Rightarrow 15 \lambda-75=0 \Rightarrow \lambda=5 \\ & \left|\begin{array}{ccc}9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}1 & 0 & 7 \\ \mu-8 & 0 & 7 \\ 16-\mu & 3 & -9\end{array}\right|=0 \\ & \Rightarrow 7-7(\mu-8)=0 \Rightarrow 1-(\mu-8)=0 \Rightarrow \mu=9 \\ & \Rightarrow \text { centre of circle }(5,9) \\ & \text { radius }=\text { length of } \perp \text { from centre }(5, \quad 9)= \\ & \left|\frac{20-27}{5}\right|=\frac{7}{5}\end{aligned}$$​​2712​333​5−2−(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​9816−μ​333​5−2−9​​=0⇒​1μ−816−μ​003​77−9​​=0⇒7−7(μ−8)=0⇒1−(μ−8)=0⇒μ=9⇒ centre of circle (5,9) radius = length of ⊥ from centre (5,9)=​520−27​​=57​​