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Question

The equation of the circle having (3,3) and origin as the endpoints of its diameter is


A

x2+y23(x+y)=0

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B

x2+y23x+y=0

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C

(x+y)22xy3(x+y)=0

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D

(x+y)2+2xy3x+y=0

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Solution

The correct options are
A

x2+y23(x+y)=0


C

(x+y)22xy3(x+y)=0


Equation of a circle when the end points (x1,y1) and (x2,y2) of its diameter are given, is given by,

(xx1)(xx2)+(yy1)(yy2)=0

So, when (3,3) and (0,0) are the end points of the diameter, equation of the circle would become:

(x3)(x0)+(y3)(y0)=0

x23x+y23y=0

x2+y23(x+y)=0

(x+y)22xy3(x+y)=0


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