Step 1: Points A aur B nikalna
Line ax+by=0 aur circle x2+y2−2x=0 ke intersection points A(α,0) aur B(1,β) hain.
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Point A(α,0) ke liye: Circle ki equation mein y=0 rakhein.
α2+0−2α=0⟹α(α−2)=0.
Toh α=0 ya α=2.
Lekin point line ax+by=0 par bhi hai, toh a(0)+b(0)=0 (hamesha sahi hai) ya a(2)+b(0)=0⟹2a=0⟹a=0 (jo valid nahi ho sakta).
Isliye, A=(0,0) aur α=0.
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Point B(1,β) ke liye: Circle mein x=1 rakhein.
12+β2−2(1)=0⟹β2−1=0⟹β=±1.
Kyuki point line ax+by=0 par hai: a(1)+b(β)=0⟹a=−bβ.
Yahan β ki value hum context ke hisaab se use karenge. Maan lete hain B=(1,1) ya (1,−1).
Step 2: Diameter AB wale circle ki equation
Agar A(0,0) aur B(1,1) diameter ke ends hain:
(x−0)(x−1)+(y−0)(y−1)=0
x2−x+y2−y=0
Iska centre C1=(1/2,1/2) aur radius square r2=1/2.
Step 3: Centre ki image nikalna
Hume is circle ki image line x+y+2=0 mein nikalni hai.
Centre (1/2,1/2) ki image (x′,y′) formula se:
1x′−1/2=1y′−1/2=−212+12(1/2+1/2+2)
1x′−1/2=1y′−1/2=−223=−3
x′=−3+0.5=−2.5=−5/2
y′=−3+0.5=−2.5=−5/2
Naya centre C2=(−5/2,−5/2).
Step 4: Nayi circle ki equation
Radius vahi rahega (r2=1/2):
x2+5x+25/4+y2+5y+25/4=1/2
Final Answer:
Sahi option (4) hai.
