The equation of the circle which cuts all the | vedclass

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(D) Let the required circle be $x^2+y^2+2gx+2fy+c=0$. Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_

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$4x^2+4y^2=7$

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(D) Let the required circle be $x^2+y^2+2gx+2fy+c=0$. Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$. The given circles are: $C_1: x^2+y^2-2x-2y+\frac{7}{4}=0$ $C_2: x^2+y^2+2x-2y+\frac{7}{4}=0$ $C_3: x^2+y^2+2x+2y+\frac{7}{4}=0$ Applying the orthogonality condition for $C_1$: $2g(-1)+2f(-1)=c+\frac{7}{4} \implies -2g-2f=c+\frac{7}{4}$ For $C_2$: $2g(1)+2f(-1)=c+\frac{7}{4} \implies 2g-2f=c+\frac{7}{4}$ For $C_3$: $2g(1)+2f(1)=c+\frac{7}{4} \implies 2g+2f=c+\frac{7}{4}$ Solving these,we get $g=0, f=0$ and $c=-\frac{7}{4}$. Thus,the equation is $x^2+y^2-\frac{7}{4}=0$,which simplifies to $4x^2+4y^2=7$.

The equation of the circle which cuts all the three circles $4(x-1)^2+4(y-1)^2=1$,$4(x+1)^2+4(y-1)^2=1$,and $4(x+1)^2+4(y+1)^2=1$ orthogonally is

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