Q . IF THE CURVES-ax2+ 4xy + 2y2 + x + y + 5 = 0 AND ax2 + 6xy + 5y2 + 2x + 3y + 8 = 0 INTERSECT AT FOUR CONCYCLIC POINTS THEN THE VALUE OF “a” IS - 4 - 46 -6

Please find below the solution to the asked query:

 

S1:ax2+4xy+2y2+x+y+5=0S2:ax2+6xy+5y2+2x+3y+8=0Equation of curve passing through their intersection will beS1+kS2=0⇒(ax2+4xy+2y2+x+y+5)+k(ax2+6xy+5y2+2x+3y+8)=0Since the points are concyclic, the above curve should be a circle , thus⇒Coefficient of xy=0⇒4+6k=0⇒k=−23Ceofficient of x2=Ceofficient of y2⇒(a+ka)=2+5k⇒a(1+k)=2+5k⇒a(1−23)=2−103⇒a(3−23)=6−103⇒a=−4 (Answer)