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Question

A point inside the circle x2+y2+3x3y+2=0 is

A
(1,3)
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B
(2,1)
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C
(2,1)
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D
(3,2)
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Solution

The correct option is B (2,1)
x2+y2+3x3y+2=0
(x+32)2+(y32)2+2=(32)2+(+32)2
(x+32)2+(y32)2=922=52=(52)2
center o (32,32) radius =52
(A)(1,3)OA=(32+1)2+(323)2
=(1/2)2+(3/2)2=102=52
OA = Radius
A Lies on the circle
(B).(2,1)OB=(2+3/2)2+(13/2)2
OB=14+14=1/2
OB< Radius
B lies inside circle

1117179_1271072_ans_4b5032c88358401b9e4257cbf5dce2dd.JPG

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