Polar of any point P w.r.t two given circles meet at Q. Prove that radical axis of circles bisect the line segment PQ

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$$\begin{gathered} \mathrm{You} \mathrm{question} \mathrm{seems} \mathrm{incomplete}. \mathrm{So} \mathrm{I}\hspace{0.17em}\mathrm{cannot} \mathrm{give} \mathrm{exact} \mathrm{solution}. \\ \mathrm{Let} \mathrm{equations} \mathrm{of} \mathrm{circle} \mathrm{be}: \\ {\mathrm{S}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}+2{\mathrm{f}}_1\mathrm{y}+{\mathrm{c}}_1=0 \\ {\mathrm{S}}_2:{\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}_2\mathrm{x}+2{\mathrm{f}}_2\mathrm{y}+{\mathrm{c}}_2=0 \\ \mathrm{Let} \mathrm{pole} \mathrm{i}.\mathrm{e}. \mathrm{point} \mathrm{P}\hspace{0.17em}\mathrm{be} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{equation} \mathrm{of} \mathrm{polar} \mathrm{is}: \\ \mathrm{T}=0 \\ \mathrm{For} {\mathrm{C}}_1 \mathrm{we} \mathrm{have}: \\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+2{\mathrm{g}}_1\left(\frac{\mathrm{x}+{\mathrm{x}}_1}{2}\right)+2{\mathrm{f}}_1\left(\frac{\mathrm{y}+{\mathrm{y}}_1}{2}\right)+{\mathrm{c}}_1=0 \\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+{\mathrm{g}}_1\left(\mathrm{x}+{\mathrm{x}}_1\right)+{\mathrm{f}}_1\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_1=0 \left(\mathrm{i}\right) \\ \mathrm{For} {\mathrm{C}}_1 \mathrm{we} \mathrm{have}: \\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+2{\mathrm{g}}_2\left(\frac{\mathrm{x}+{\mathrm{x}}_1}{2}\right)+2{\mathrm{f}}_2\left(\frac{\mathrm{y}+{\mathrm{y}}_1}{2}\right)+{\mathrm{c}}_2=0 \\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+{\mathrm{g}}_2\left(\mathrm{x}+{\mathrm{x}}_1\right)+{\mathrm{f}}_2\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_2=0 \left(\mathrm{ii}\right) \\ \mathrm{Equate} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+{\mathrm{g}}_1\left(\mathrm{x}+{\mathrm{x}}_1\right)+{\mathrm{f}}_1\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_1={\mathrm{xx}}_1+{\mathrm{yy}}_1+{\mathrm{g}}_2\left(\mathrm{x}+{\mathrm{x}}_1\right)+{\mathrm{f}}_2\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_2 \\ \Rightarrow {\mathrm{g}}_1+{\mathrm{f}}_1\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_1={\mathrm{g}}_2\left(\mathrm{x}+{\mathrm{x}}_1\right)+{\mathrm{f}}_2\left(\mathrm{y}+{\mathrm{y}}_1\right)+{\mathrm{c}}_2 \\ \Rightarrow \left(\mathrm{x}+{\mathrm{x}}_1\right)\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+\left(\mathrm{y}+{\mathrm{y}}_1\right)\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{c}}_1-{\mathrm{c}}_2=0 \\ \Rightarrow \mathrm{x}\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+{\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+\mathrm{y}\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{c}}_1-{\mathrm{c}}_2=0 \\ \Rightarrow \mathrm{x}\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+\mathrm{y}\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{c}}_1-{\mathrm{c}}_2=0 \\ \Rightarrow 2\mathrm{x}\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+2\mathrm{y}\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+2{\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+2{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+2{\mathrm{c}}_1-2{\mathrm{c}}_2=0 ;\left(\mathrm{iii}\right) \\ \mathrm{Radical} \mathrm{axis} \mathrm{of} \mathrm{two} \mathrm{circles} \mathrm{will} \mathrm{be}: \\ 2\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)\mathrm{x}+2\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)\mathrm{y}+{\mathrm{c}}_1-{\mathrm{c}}_2=0 ;\left(\mathrm{iv}\right) \\ \mathrm{By} \left(\mathrm{iii}\right) \mathrm{and} \left(\mathrm{iv}\right), \mathrm{we} \mathrm{get}: \\ {\mathrm{c}}_2-{\mathrm{c}}_1+2{\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+2{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+2{\mathrm{c}}_1-2{\mathrm{c}}_2=0 \\ \Rightarrow 2{\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+2{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+{\mathrm{c}}_1-{\mathrm{c}}_2=0 \\ \Rightarrow {\mathrm{x}}_1\left({\mathrm{g}}_1-{\mathrm{g}}_2\right)+{\mathrm{y}}_1\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)+\frac{{\mathrm{c}}_1-{\mathrm{c}}_2}{2}=0 \end{gathered}$$

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