Question

Coordinates of the centre of the circle which bisects the circumferences of the circles $x^2+y^2=1:x^2+y^2+2x-3=0$ and $x^2+y^2+2y-3=0$ is:

• A. $(-3,-3)$
• B. $(3,3)$
• C. $(2,2)$
• D. $(-2,-2)$

Answer 1

The correct option is A $(-3,-3)$

Let the equation of circle bisects the $3$ circles $x^2+y^2=1$

$\begin{array}{c}{x}^2+y^2+2x-3=0\\ x^2+y^2+2y-3=0\end{array}$

$x^2+y^2+2gx+2fy+c=0$

$=gx+fy+c-1=0$

Passes throw $(0,0)$ i.e.

Center of $x^2+y^2=1$

$C=1$

$2gx+fy+c-2x+3=0$

Pass throw $(-1,0)$ i.e. center of $x^2+y^2+2x-3=0$

$\begin{array}{c}-2g+c+5=0\Rightarrow -2g+6=0\\ \Rightarrow g=3\\ 2gx+2fy+c-2y+3=0\end{array}$

Pass throw $(0,-1)$

$\begin{array}{c}-2f+c+5=0\\ f=3\\ x^2+y^2+6x+6y+1=0\\ C=\left(-3,-3\right)\end{array}$

$\therefore$ Option $A$ is correct answer.