Question

The equation of the circle which passes through the origin and cuts off intercepts of 2 units length from negative coordinate axes, is.

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

Answer 1

Answer: (C)

$x ^ { 2 } + y ^ { 2 } + 2 x + 2 y = 0$

Answer 2

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).