UPSC 2015 CDS I Math Paper

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