JEE Main 2013 — Circle Question with Solution

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 $=(1,1)$ $$\therefore a^2-7a+11=1\Rightarrow a=2,5$$ and $a^2-6a+6=1\Rightarrow a=1,5$ From both (i) and (ii), $a=5$ Now on replacing each of $\left(a^2-7a+11\right)$ and $\left(a^2-6a+6\right)$ by 1 , the equation of the given circle is $x^2+y^2-2x-2y+b^3+1=0$ $$\begin{array}{rl}& \Rightarrow (x-1{)}^2+(y-1{)}^2+b^3=1\\ & \Rightarrow b^3=1-\left[(x-1{)}^2+(y-1{)}^2\right]\\ & \therefore b\in (-\infty ,1)\end{array}$$