If each of the lines 5 x+8 y=13 and 4 x-y=3 contains a diameter of the circle x^2+y^2-2 (a^2-7 a+11 )x-2 (a^2-6 a+6 ) y+b^3+1=0 , then :

Step-by-step 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) a^2-7 a+11=1 a=2,5 and a^2-6 a+6=1 a=1,5 From both (i) and (ii), a=5 Now on replacing each of (a^2-7 a+11 ) and (a^2-6 a+6 ) by 1 , the equation of the given circle is x^2+y^2-2 x-2 y+b^3+1=0 aligned & (x-1)^2+(y-1)^2+b^3=1 & b^3=1- [(x-1)^2+(y-1)^2 ] & b (- , 1) aligned