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

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{circle} \mathrm{having} (0,-1)(-2,3) \mathrm{are} \mathrm{end} \mathrm{points} \mathrm{of} \mathrm{diameter} is \\ (\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)+(\mathrm{y}-{\mathrm{y}}_1)(\mathrm{y}-{\mathrm{y}}_2)=0 \\ (\mathrm{x}+0)(\mathrm{x}+2)+(\mathrm{y}+1)(\mathrm{y}-3)=0 \\ {\mathrm{x}}^2+2\mathrm{x}+{\mathrm{y}}^2-2\mathrm{y}-3=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}-3=0 — \left(1\right) \\ \mathrm{The} \mathrm{required} \mathrm{circle} \mathrm{touching} \mathrm{the} \mathrm{line} 2\mathrm{x}+3\mathrm{y}+1=0 \\ \mathrm{Then} \mathrm{r}=\mathrm{d} — \left(2\right) \\ \mathrm{By} \mathrm{verification} fourth option \\ \mathrm{The} \mathrm{equation} \mathrm{of} \mathrm{required} \mathrm{circle} \mathrm{is} 2x^2+2y^2-10x-5y+1=0 which satiesfies \\ the given conditions \end{gathered}$$