Number of circles intersecting x^2+y^2=4,x^2+ | vedclass

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(A) Let the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. For a circle to be orthogonal to $x^2+y^2+2g_i x+2f_i y+c_i=0$,the condition is $2gg_i+

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$0$

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(A) Let the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. For a circle to be orthogonal to $x^2+y^2+2g_i x+2f_i y+c_i=0$,the condition is $2gg_i+2ff_i=c+c_i$. Given circles: $S_1: x^2+y^2-4=0 \Rightarrow g_1=0, f_1=0, c_1=-4$ $S_2: x^2+y^2-2x-3=0 \Rightarrow g_2=-1, f_2=0, c_2=-3$ $S_3: x^2+y^2-2y-3=0 \Rightarrow g_3=0, f_3=-1, c_3=-3$ Applying the condition: $1$) For $S_1$: $2g(0)+2f(0)=c-4 \Rightarrow c=4$. $2$) For $S_2$: $2g(-1)+2f(0)=c-3$ $\Rightarrow -2g=4-3=1$ $\Rightarrow g=-1/2$. $3$) For $S_3$: $2g(0)+2f(-1)=c-3$ $\Rightarrow -2f=4-3=1$ $\Rightarrow f=-1/2$. Now,check the radius $r$ of the circle $S$: $r^2 = g^2+f^2-c = (-1/2)^2+(-1/2)^2-4 = 1/4+1/4-4 = 1/2-4 = -7/2$. Since $r^2 Thus,the number of such circles is $0$.

Number of circles intersecting $x^2+y^2=4$,$x^2+y^2-2x-3=0$ and $x^2+y^2-2y-3=0$ orthogonally is

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