Point of intersection of two given lines is (1, 1). Sinceeach of the two given lines contains a diameter of thegiven circle, therefore the point of intersection of the twogiven lines is the centre of the given circle.Hence centre = (1, 1) ∴a2−7a+11=1⇒a=2,5 and a2−6a+6=1⇒a=1,5 From both (i) and (ii), a=5 Now on replacing each of (a2−7a+11) and (a2−6a+6) by 1, the equation of the given circle is x2+y2−2x−2y+b3+1=0 ⇒(x−1)2+(y−1)2+b3=1 ⇒b3=1−[(x−1)2+(y−1)2] ∴b∈(−∞,1)