If the radical centre of the three circles x^ | vedclass

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(D) Let the equations of the circles be $S_1: x^2+y^2-1=0$,$S_2: x^2+y^2-2x-3=0$,and $S_3: x^2+y^2-2y-3=0$. To find the radical centre,we find the int

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$(x+1)^2+(y+1)^2=25$

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(D) Let the equations of the circles be $S_1: x^2+y^2-1=0$,$S_2: x^2+y^2-2x-3=0$,and $S_3: x^2+y^2-2y-3=0$. To find the radical centre,we find the intersection of the radical axes. $S_1-S_2=0 \implies (x^2+y^2-1) - (x^2+y^2-2x-3) = 0 \implies 2x+2=0 \implies x=-1$. $S_1-S_3=0 \implies (x^2+y^2-1) - (x^2+y^2-2y-3) = 0 \implies 2y+2=0 \implies y=-1$. Thus,the radical centre $C(\alpha, \beta)$ is $(-1, -1)$. The radii of the circles are: $r_1 = \sqrt{0^2+0^2-(-1)} = 1$. $r_2 = \sqrt{1^2+0^2-(-3)} = \sqrt{4} = 2$. $r_3 = \sqrt{0^2+1^2-(-3)} = \sqrt{4} = 2$. The sum of the radii is $r = 1+2+2 = 5$. The equation of the circle with centre $(-1, -1)$ and radius $5$ is $(x-(-1))^2 + (y-(-1))^2 = 5^2$,which simplifies to $(x+1)^2+(y+1)^2=25$.

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

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