If each of the lines 5 x+8 y=13 and 4 x-y=3 contains a diameter of...

Point of intersection of two given lines is $(1,1)$. Since each of the two given lines contains a diameter of the given circle, therefore the point of intersection of the two given lines is the centre of the given circle. Hence centre $=(1,1)$ \therefore a²-7 a+11=1 \Rightarrow a=2,5 and $a^2-6 a+6=1 \Rightarrow a=1,5$ From both (i) and (ii), $a=5$ Now on replacing each of $\left(a^2-7 a+11\right)$ and $\left(a^2-6 a+6\right)$ by 1 , the equation of the given circle is $x^2+y^2-2 x-2 y+b^3+1=0$$\begin{aligned}& \Rightarrow(x-1)^2+(y-1)^2+b³=1 \\ & \Rightarrow b³=1-\left[(x-1)^2+(y-1)^2\right] \\ & \therefore b \in(-\infty,1) \end{aligned}$$