A circle S cuts three circles x^2+y^2-4x-2y+4 | vedclass

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(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. F

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$\sqrt{\frac{29}{8}}$

Vedclass · Answer

(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. For $x^2+y^2-4x-2y+4=0$: $2g(-2) + 2f(-1) = c+4 \Rightarrow -4g-2f = c+4$ $(i)$. For $x^2+y^2-2x-4y+1=0$: $2g(-1) + 2f(-2) = c+1 \Rightarrow -2g-4f = c+1$ $(ii)$. For $x^2+y^2+4x+2y+1=0$: $2g(2) + 2f(1) = c+1 \Rightarrow 4g+2f = c+1$ $(iii)$. Adding $(i)$ and $(iii)$,we get $0 = 2c+5$,so $c = -\frac{5}{2}$. Substituting $c$ into $(iii)$: $4g+2f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 8g+4f = -3$ $(iv)$. Substituting $c$ into $(ii)$: $-2g-4f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 4g+8f = 3$ $(v)$. Adding $(iv)$ and $(v)$: $12g+12f = 0 \Rightarrow g = -f$. Substituting $g = -f$ into $(iv)$: $-8f+4f = -3$ $\Rightarrow -4f = -3$ $\Rightarrow f = \frac{3}{4}, g = -\frac{3}{4}$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{3}{4})^2 + (\frac{3}{4})^2 - (-\frac{5}{2})} = \sqrt{\frac{9}{16} + \frac{9}{16} + \frac{40}{16}} = \sqrt{\frac{58}{16}} = \sqrt{\frac{29}{8}}$.

$A$ circle $S$ cuts three circles $x^2+y^2-4x-2y+4=0$,$x^2+y^2-2x-4y+1=0$,and $x^2+y^2+4x+2y+1=0$ orthogonally. Then,the radius of $S$ is

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