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Question

x2+y2+2x4y+k=0 passes through

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Solution

A) For k=0
x2+y2+2x4y+k=0x2+y2+2x4y=0x2+y2+2x4y+1+414(x+1)2+(y2)2=5
Point (0,0) and (0,4) satisfies this
B) For k=1
x2+y2+2x4y+k=0x2+y2+2x4y+1=0x2+y2+2x4y+1+44(x+1)2+(y2)2=4
Point (1,2) satisfies this
C) For k=15
x2+y2+2x4y+k=0x2+y2+2x4y15=0x2+y2+2x4y+1+420(x+1)2+(y2)2=20
Point (3,4) satisfies this
D) For k=2
x2+y2+2x4y+k=0x2+y2+2x4y1=0x2+y2+2x4y+1+46(x+1)2+(y2)2=2
Point (0, 1) satisfies

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