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

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$\left(-\frac{19}{52}, \frac{55}{52}\right)$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through $(1,1)$,we have $1+1+2g+2f+c=0$,which simplifies to $2g+2f+c+2=0$ $(i)$. Since the circle is orthogonal to $x^2+y^2+3x-5y+7=0$,we have $2g(3/2) + 2f(-5/2) = c+7$,which simplifies to $3g-5f-c-7=0$ (ii). Since the circle is orthogonal to $x^2+y^2-6x-10y+9=0$,we have $2g(-3) + 2f(-5) = c+9$,which simplifies to $6g+10f+c+9=0$ (iii). Adding $(i)$ and (ii): $(2g+2f+c+2) + (3g-5f-c-7) = 0 \Rightarrow 5g-3f-5=0$ (iv). Adding $(i)$ and (iii): $(2g+2f+c+2) + (6g+10f+c+9) = 0 \Rightarrow 8g+12f+2c+11=0$. From $(i)$,$c = -2g-2f-2$. Substituting this into (iii): $6g+10f+(-2g-2f-2)+9=0 \Rightarrow 4g+8f+7=0$ $(v)$. Solving (iv) and $(v)$: From (iv),$f = (5g-5)/3$. Substituting into $(v)$: $4g + 8((5g-5)/3) + 7 = 0$ $\Rightarrow 12g + 40g - 40 + 21 = 0$ $\Rightarrow 52g = 19$ $\Rightarrow g = 19/52$. Then $f = (5(19/52)-5)/3 = (95/52 - 260/52)/3 = (-165/52)/3 = -55/52$. The centre is $(-g, -f) = (-19/52, 55/52)$. Thus,option $(d)$ is correct.

The centre of the circle passing through the point $(1,1)$ and orthogonal to the circles $x^2+y^2+3x-5y+7=0$ and $x^2+y^2-6x-10y+9=0$ is

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