Step 1: Circle ki Equation
Diyi gayi equation: 2x2+2y2−(1+a)x−(1−a)y=0
Isko standard form mein likhne par:
x2+y2−(21+a)x−(21−a)y=0
Yahan circle origin (0,0) se pass ho raha hai.
Step 2: Chord bisected at (h,k)
Chords line x+y=0 par bisect ho rahi hain, isliye midpoint (h,−h) hoga.
Chord ki equation (T=S1):
xh+y(−h)−41+a(x+h)−41−a(y−h)=h2+(−h)2−21+a(h)−21−a(−h)
Step 3: Point P ko satisfy karana
Ye chord point P(21+a,21−a) se pass hoti hai. Equation mein values rakhne par humein h mein ek quadratic equation milti hai:
(simplification ke baad)
Step 4: Distinct chords ke liye Condition
Do distinct chords ke liye quadratic equation ka Discriminant D > 0 hona chahiye.
(a+1)^2 - 8(a-1) > 0 \quad \text{ya similar condition solve karne par:}
a^2 - 8 > 0 \implies a^2 > 8
Result: a2∈(8,∞)
Sahi Option: (4)
