x2−y2=0....(1) (x−a)2+y2=1....(2) Adding both the equations, we get: (x−a)2+x2=1⇒2x2+a2−2ax =1⇒2x2−2ax+(a2−1)=0 Since we have only one root, therefore, discriminant = 0. (−2a)2−4×2×(a2−1)=0 ⇒4a2−8a2+8=0 ⇒4a2=8⇒a2=2 ⇒a=±√2,but it is given that a≠−√2∴a=√2