If the circle S equiv x^2+y^2-4=0 intersects | vedclass

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(B) Given $S \equiv x^2+y^2-4=0$,so $C_1(0,0)$ and $r_1=2$.Let the center of $S^{\prime}=0$ be $C_2(h, k)$ and radius $r_2=\frac{5 \sqrt{2}}{2}$.The e

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$\left(-\frac{\sqrt{2}}{2}, 2 \sqrt{2}\right)$ or $\left(\frac{\sqrt{2}}{2},-2 \sqrt{2}\right)$

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(B) Given $S \equiv x^2+y^2-4=0$,so $C_1(0,0)$ and $r_1=2$.Let the center of $S^{\prime}=0$ be $C_2(h, k)$ and radius $r_2=\frac{5 \sqrt{2}}{2}$.The equation of $S^{\prime}=0$ is $(x-h)^2+(y-k)^2=\left(\frac{5 \sqrt{2}}{2}\right)^2 = \frac{25}{2}$.Expanding this,we get $x^2+y^2-2xh-2yk+h^2+k^2-\frac{25}{2}=0$.The equation of the common chord is $S-S^{\prime}=0$,which is $2xh+2yk-(h^2+k^2-\frac{25}{2})-4=0$,or $2xh+2yk = h^2+k^2-\frac{17}{2}$.The slope of the common chord is $-\frac{2h}{2k} = -\frac{h}{k}$. Given the slope is $\frac{1}{4}$,we have $-\frac{h}{k} = \frac{1}{4} \Rightarrow k = -4h$.The common chord has maximum length when it passes through the center of the smaller circle $C_1(0,0)$.Substituting $(0,0)$ into the equation of the chord: $2(0)h+2(0)k = h^2+k^2-\frac{17}{2} \Rightarrow h^2+k^2 = \frac{17}{2}$.Substituting $k=-4h$ into $h^2+k^2=\frac{17}{2}$,we get $h^2+(-4h)^2 = \frac{17}{2}$ $\Rightarrow 17h^2 = \frac{17}{2}$ $\Rightarrow h^2 = \frac{1}{2}$ $\Rightarrow h = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.If $h = \frac{\sqrt{2}}{2}$,then $k = -4(\frac{\sqrt{2}}{2}) = -2\sqrt{2}$.If $h = -\frac{\sqrt{2}}{2}$,then $k = -4(-\frac{\sqrt{2}}{2}) = 2\sqrt{2}$.Thus,the center $C_2$ is $\left(\frac{\sqrt{2}}{2}, -2\sqrt{2}\right)$ or $\left(-\frac{\sqrt{2}}{2}, 2\sqrt{2}\right)$.

If the circle $S \equiv x^2+y^2-4=0$ intersects another circle $S^{\prime}=0$ of radius $\frac{5 \sqrt{2}}{2}$ in such a manner that the common chord is of maximum length with slope equal to $\frac{1}{4}$,then the centre of $S^{\prime}=0$ is

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