Question

The equation of the circle describes on the common chord of the circles $x^2+y^2+2x=0$ and $x^2+y^2+2y=0$ as diameter is

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

Answer 1

Answer: (C)

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

$s_1:x^2+y^2+2x=0$
Centre $\equiv (-1,0),$ radius=1
$s^2:x^2+y^2+2y=0$
Centre $\equiv (0,-1),$ radius =1
$P$ is the mid-point of $C_1$ and $C_2$ becomes radius of both circles are equal.
$\therefore P\equiv (\frac{-1}{2},\frac{-1}{2})$
And $AP=\sqrt{c_1{A}^2-c_1{p}^2}=\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}$
$AP=\frac{1}{\sqrt{2}}$
$\therefore$ Equation of circle having $AB$ as diameter is
${(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2=\frac{1}{2}$
$\Rightarrow x^2+y^2+x+y=0$