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(D) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. Since it passes through $(1,0)$,we have $1^2+0^2+2g(1)+2f(0)+c=0$,which gives $2

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$(0,0)$

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(D) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. Since it passes through $(1,0)$,we have $1^2+0^2+2g(1)+2f(0)+c=0$,which gives $2g+c=-1$ (Equation $1$). The circle is orthogonal to $x^2+y^2-2x+4y+1=0$. The condition for orthogonality $2g_1g_2+2f_1f_2=c_1+c_2$ gives $2g(-1)+2f(2)=c+1$,so $-2g+4f=c+1$ (Equation $2$). The circle is also orthogonal to $x^2+y^2+6x-2y+1=0$. This gives $2g(3)+2f(-1)=c+1$,so $6g-2f=c+1$ (Equation $3$). Subtracting Equation $2$ from Equation $3$: $(6g-2f) - (-2g+4f) = (c+1) - (c+1)$,which simplifies to $8g-6f=0$,or $f = \frac{4}{3}g$. Substituting $f = \frac{4}{3}g$ into Equation $2$: $-2g+4(\frac{4}{3}g) = c+1$,so $-2g+\frac{16}{3}g = c+1$,which gives $\frac{10}{3}g = c+1$,or $c = \frac{10}{3}g-1$. Substituting $c = \frac{10}{3}g-1$ into Equation $1$: $2g + (\frac{10}{3}g-1) = -1$,so $\frac{16}{3}g = 0$,which means $g=0$. Then $f = \frac{4}{3}(0) = 0$ and $c = \frac{10}{3}(0)-1 = -1$. The center of the circle is $(-g, -f) = (0,0)$.

The center of the circle passing through the point $(1,0)$ and cutting the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2+6x-2y+1=0$ orthogonally is

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