If the common chord of the circles x^2+y^2+4y | vedclass

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(B) The equations of the circles are $S_1: x^2+y^2+4y=0$ and $S_2: x^2+y^2-4x-5=0$.The equation of the common chord is given by $S_1 - S_2 = 0$,which

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$\frac{3}{8}$

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(B) The equations of the circles are $S_1: x^2+y^2+4y=0$ and $S_2: x^2+y^2-4x-5=0$.The equation of the common chord is given by $S_1 - S_2 = 0$,which is $(x^2+y^2+4y) - (x^2+y^2-4x-5) = 0$,simplifying to $4x+4y+5=0$.Let the circle $S=0$ have its centre at $(h, k)$. Since the common chord is the diameter of $S=0$,the centre $(h, k)$ must lie on the common chord $4x+4y+5=0$. Thus,$4h+4k+5=0$.Also,the centre of $S=0$ must be the midpoint of the common chord if the circle is uniquely determined by the chord as its diameter. The common chord intersects the line joining the centres of $S_1$ and $S_2$. The centre of $S_1$ is $C_1(0, -2)$ and the centre of $S_2$ is $C_2(2, 0)$.The line joining the centres $C_1$ and $C_2$ has the equation $y - 0 = \frac{-2-0}{0-2}(x-2)$,which simplifies to $y = x-2$ or $x-y-2=0$.The centre $(h, k)$ of the circle $S=0$ is the intersection of the common chord $4x+4y+5=0$ and the line of centres $x-y-2=0$.Solving the system:$4x+4y = -5$$x-y = 2 \Rightarrow y = x-2$Substituting $y$ in the first equation: $4x + 4(x-2) = -5$ $\Rightarrow 8x - 8 = -5$ $\Rightarrow 8x = 3$ $\Rightarrow x = \frac{3}{8}$.Thus,the abscissa of the centre is $\frac{3}{8}$.

If the common chord of the circles $x^2+y^2+4y=0$ and $x^2+y^2-4x-5=0$ is the diameter of the circle $S=0$,then the abscissa of the centre of the circle $S=0$ is

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