Answers

2014-05-20T18:46:05+05:30
X²+y²-2x-4y+1 = 0  --------------------(1)
2x+2yy'-2-4y' = 0                 (differentiating)
y' = (1-x)/(y-2)
since tangent is parallel to X axis , y' = 0
1-x = 0 
x = 1
put x = 1 in equation (1)
1+y²-2-4y+1 = 0
y²-4y = 0
y = 0   ,  4
hence points = (1,0)&(1,4)
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