x2−y2=0..........(1) (x−k)2+y2=1.....(2) Adding the given system of equations, we have (x−k)2+x2=1 ⇒2x2−2kx+k2−1=0 The given system of equations yields a unique positive solution of x if D=(−2k)2−4×2×(k2−1)=0 ⇒−4k2+8=0 ⇒k2=2 ⇒k=√2 (For k=−√2 we get negative value of x) Alternative Method: x2−y2=0⇒x=y ∴(x−k)2+y2=1 ⇒(x−k)2+x2=1 ⇒2x2−2kx+k2−1=0 yields a unique solution if −(2k)2−4×2×(k2−1)=0 ⇒−4k2=−8 ⇒−k=√2