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

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$2x^2+2y^2-10x-5y+1=0$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. Since it passes through $(1,-1)$,we have $1+1+2g-2f+c=0$,which simplifies to $2g-2f+c=-2$ $(i)$. The normal to the circle at $(1,-1)$ is perpendicular to the tangent $2x+3y+1=0$. The slope of the tangent is $-2/3$,so the slope of the normal is $3/2$. The equation of the normal is $y-(-1) = \frac{3}{2}(x-1)$,which simplifies to $3x-2y-5=0$. The center $(-g, -f)$ lies on the normal,so $3(-g)-2(-f)-5=0$,or $-3g+2f=5$ (ii). The circle is orthogonal to the circle with diameter endpoints $(0,-1)$ and $(-2,3)$,which is $x(x+2) + (y+1)(y-3) = 0$,or $x^2+y^2+2x-2y-3=0$. The condition for orthogonality $2g_1g_2 + 2f_1f_2 = c_1+c_2$ gives $2g(1) + 2f(-1) = c-3$,or $2g-2f-c=-3$ (iii). Adding $(i)$ and (iii): $(2g-2f+c) + (2g-2f-c) = -2-3$,so $4g-4f=-5$. Solving the system of equations,we find $g=-5/2, f=-5/4, c=1/2$. Substituting these into the circle equation: $x^2+y^2-5x-\frac{5}{2}y+\frac{1}{2}=0$,which is $2x^2+2y^2-10x-5y+1=0$.

The equation of the circle touching the line $2x+3y+1=0$ at the point $(1,-1)$ and orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as diameter,is

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