Question

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

• A. $x^2+y^2-3(x+y)=0$
• B. $x^2+y^2-3x+y=0$
• C. $(x+y{)}^2-2xy-3(x+y)=0$
• D. $(x+y{)}^2+2xy-3x+y=0$

Answer 1

Answer: (A), (C)

The correct options are
A

$x^2+y^2-3(x+y)=0$

C

$(x+y{)}^2-2xy-3(x+y)=0$

Equation of a circle when the end points $(x_1,y_1)$ and $(x_2,y_2)$ of its diameter are given, is given by,

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

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

$(x-3)(x-0)+(y-3)(y-0)=0$

$⟹$ $x^2-3x+y^2-3y=0$

$⟹$ $x^2+y^2-3(x+y)=0$

$⟹$ $(x+y{)}^2-2xy-3(x+y)=0$