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

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$$\begin{gathered} \mathrm{Diametric} \mathrm{end} \mathrm{of} \mathrm{given} \mathrm{circle} \mathrm{are} \left(0,3\right) \mathrm{and} \left(-2,-1\right). \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{cirlce} \mathrm{is} \\ \left(\mathrm{x}-0\right)\left(\mathrm{x}+2\right)+\left(\mathrm{y}-3\right)\left(\mathrm{y}+1\right)=0 \\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}+{\mathrm{y}}^2-2\mathrm{y}-3=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}-3=0 \\ \mathrm{Let} \mathrm{equation} \mathrm{of} \mathrm{required} \mathrm{circle} \mathrm{be} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ \mathrm{Applying} \mathrm{condition} \mathrm{of} \mathrm{orthogonality} \mathrm{we} \mathrm{get}: \\ 2\mathrm{g}\left(1\right)+2\mathrm{f}\left(-1\right)=\mathrm{c}-3 \\ \Rightarrow 2\mathrm{g}-2\mathrm{f}=\mathrm{c}-3....\left(\mathrm{i}\right) \\ \left(1,-1\right) \mathrm{lies} \mathrm{on} \mathrm{curve} \\ 1+1+2\mathrm{g}-2\mathrm{f}+\mathrm{c}=0 \\ \Rightarrow 2\mathrm{g}-2\mathrm{f}=-\mathrm{c}-2....\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{c}-3=-\mathrm{c}-2 \\ \Rightarrow \mathrm{c}=\frac{1}{2} \\ \Rightarrow \left(\mathrm{i}\right) \mathrm{becomes} \\ 2\mathrm{g}-2\mathrm{f}=\frac{1}{2}-3=-\frac{5}{2} \\ \Rightarrow \mathrm{g}-\mathrm{f}=-\frac{5}{4}...\left(\mathrm{iii}\right) \\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\frac{1}{2}=0 \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2+4\mathrm{gx}+4\mathrm{fy}+1=0 \\ \mathrm{Now} \mathrm{as} \mathrm{curve} \mathrm{and} \mathrm{line} \mathrm{touch} \mathrm{each} \mathrm{other} \mathrm{at} \left(1,-1\right) \mathrm{hence} \mathrm{both} \mathrm{will} \mathrm{hav} \\ \mathrm{same} \mathrm{slope} \mathrm{at} \mathrm{that} \mathrm{point}. \\ \mathrm{Slope} \mathrm{of} \mathrm{line} 2\mathrm{x}+3\mathrm{y}+1=0 \mathrm{is} -\frac{2}{3} \\ \mathrm{Now} \\ 2{\mathrm{x}}^2+2{\mathrm{y}}^2+4\mathrm{gx}+4\mathrm{fy}+1=0 \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ 4\mathrm{x}+4\mathrm{y}.\frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{g}+4\mathrm{f}.\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \mathrm{x}+\mathrm{y}.\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{g}+\mathrm{f}.\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \mathrm{At} \left(1,-1\right), \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2}{3} \\ 1-1\left(-\frac{2}{3}\right)+\mathrm{g}+\mathrm{f}\left(-\frac{2}{3}\right)=0 \\ \Rightarrow 1+\frac{2}{3}+\mathrm{g}-\frac{2}{3}\mathrm{f}=0 \\ \mathrm{Use} \left(\mathrm{iii}\right) \\ 1+\frac{2}{3}+\mathrm{f}-\frac{5}{4}-\frac{2}{3}\mathrm{f}=0 \\ \Rightarrow \frac{\mathrm{f}}{3}=\frac{5}{4}-\frac{5}{3} \\ \Rightarrow \frac{\mathrm{f}}{3}=\left(\frac{-5}{12}\right) \\ \Rightarrow \mathrm{f}=-\frac{5}{4} \\ \mathrm{Put} \mathrm{in} \left(\mathrm{ii}\right) \mathrm{to} \mathrm{get} \\ \mathrm{g}=-\frac{5}{2} \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is} \\ 2{\mathrm{x}}^2+2{\mathrm{y}}^2-10\mathrm{x}-5\mathrm{y}+1=0 \end{gathered}$$

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