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If the radical centre of the given three circles $x^2+y^2=1, x^2+y^2-2 x-3=0$ and $x^2+y^2-2 y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles, then the circle with $C(\alpha, \beta)$ as centre and r as radius is
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Verified Answer
The correct answer is:
$(x+1)^2+(y+1)^2=25$
$\mathrm{S}_1: x^2+y^2-1=0 ; \mathrm{S}_2: x^2+y^2-2 x-3=0$
$\mathrm{S}_3: x^2+y^2-2 y-3=0$
$S_1-S_2=0$(Equation of radical axis)
$$
\begin{aligned}
& 2 x+2=0 \\
& x=-1 \\
& \mathrm{~S}_1-\mathrm{S}_3=0 \\
& 2 y+2=0 \\
& y=-1
\end{aligned}
$$
$\therefore$ Radical centre is $(-1,-1)$
$$
r_1=1, r_2=2, r_3=2
$$
$\therefore$ Required circle have radius $=5$
$\therefore$ Equation of circle : $(x+1)^2=(y+1)^2=25$
$\mathrm{S}_3: x^2+y^2-2 y-3=0$
$S_1-S_2=0$(Equation of radical axis)
$$
\begin{aligned}
& 2 x+2=0 \\
& x=-1 \\
& \mathrm{~S}_1-\mathrm{S}_3=0 \\
& 2 y+2=0 \\
& y=-1
\end{aligned}
$$
$\therefore$ Radical centre is $(-1,-1)$
$$
r_1=1, r_2=2, r_3=2
$$
$\therefore$ Required circle have radius $=5$
$\therefore$ Equation of circle : $(x+1)^2=(y+1)^2=25$
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