Coordinates of the centre of the circle which | vedclass

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Question

If $S=0$ bisects the circumference of $S^{\prime}=0$ then the common chord will pas thoo ufh the $ 	ext { centre of } S^{\prime}=0 	ext { . } $

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Vedclass · Accepted Answer

$(- 2, - 2)$

Vedclass · Answer

If $S=0$ bisects the circumference of $S^{\prime}=0$ then the common chord will pas thoo ufh the $ 	ext { centre of } S^{\prime}=0 	ext { . } $

$S=0$

$x^{2}+y^{2}+2 g x+2 f y+c=0$

$x^{2}+y^{2}-1=0$

$c \equiv(0,0)$

$S=0 \quad$ and $\quad S_{1}=0$

Common chord $\Rightarrow \quad S-S_{1}=0$

$\left(x^{2}+y^{2}+2 g x+2 f y+cight)-\left(x^{2}+y^{2}-1ight)=0$

$2 g x+2 f y+c+1=0$

$\Rightarrow C+1=0$

$\Rightarrow C=-1$

$S=0$

$x^{2}+y^{2}+2 g x+2 f y-1=0 \quad, \quad(-9,-f)$

$x^{2}+y^{2}+2 x-3=0$

$c^{\prime} \equiv(-1,0)$

$Eq ^{n} $ common $\quad$ chord $\quad \Rightarrow \quad\left(z^{2}+y^{k}+2 g x+2 f y-1ight)-\left(z^{2}+y^{2}+2 x-3ight)=0$

$\Rightarrow \quad 2 g x+2 f y-1-2 x+3=0$

$\Rightarrow \quad-2 g+0-1-2(-1)+3=0$

$\overrightarrow{7} \quad-2 g-1+2+3=0$

$\Rightarrow \quad-2 g=-4$

$\overrightarrow{7} \quad g=2$

$S=0$

$x^{2}+y^{2}+4 x+2 f y+(-1)=0$

$x^{2}+y^{2}+2 y-3=0$

$c^{\prime \prime} \equiv(0,-1)$

$Eq ^{n}  common \quad$ chord $\quad \Rightarrow \quad\left(x^{2}+y^{2}+4 x+2 f y-1ight)-\left(x^{2}+y^{2}+2 y-3ight)=0$

$\Rightarrow \quad 4 x+2 f y-1-2 y+3=0$

$\Rightarrow \quad 0 \quad-2 f-1+2+3=0$

$\Rightarrow \quad-2 f=-4$

$\Rightarrow \quad i \overline{f=2}angle$

(-2,-2) is the centre of the requised circle

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

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