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The equation of the circle which passes through the origin and cuts off intercepts of 2 units length from negative coordinate axes, is
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The correct answer is:
$x^2+y^2+2 x+2 y=0$
Since the circle passes through $(0,0)$, hence $c=0$. Also $2 \sqrt{g^2-c}=2 \Rightarrow g=1$ and $2 \sqrt{f^2-c}=2 \Rightarrow f=1$
Hence radius is $\sqrt{2}$ and centre is $(-1,-1)$. Therefore, the required equation is $x^2+y^2+2 x+2 y=0$
Trick : Obviously the centre of circle lies in III quadrant, which is given by (3).
Hence radius is $\sqrt{2}$ and centre is $(-1,-1)$. Therefore, the required equation is $x^2+y^2+2 x+2 y=0$
Trick : Obviously the centre of circle lies in III quadrant, which is given by (3).
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