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Question : 127 of 200
Marks:
+1,
-0
Solution:
Equation of circles are
‌x2+y2+2x−2y+2=0 . . . (i)
‌‌ and ‌x2+y2−‌x−‌y+‌=0 . . . (ii)
‌⇒‌‌(x2+y2+2x−2y+2)‌+λ[x2+y2−‌x−‌y+‌]=0‌⇒(1+λ)x2+(1+λ)y2+2(1−‌)x‌−2(1+‌)y+(2+‌)=0‌⇒‌‌x2+y2+2‌x‌−‌y+‌=0 . . . (iii)
‌‌ At ‌‌‌g=0⇒‌=0⇒λ=5‌, ‌ Then
x2+y2−3y+‌=0∴ Centre is
(0,‌).
The limiting points are
(−1,1),(‌,‌).
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