The correct answer is (C):
Let the 6 cm long chord be x cm away from the centre of the circle. Let the radius of the circle be r cm.
The perpendiculars from the centre of the circle to the chords bisect the chords.
r2=x2+32=(x+1)2+22
Solving, x=2 and r=√13
The equation of a circle which touches the straight lines $x + y = 2$, $x - y = 2$ and also touches the circle $x^2 + y^2 = 1$ is: