The condition that the two tangents to the parabola y²=4ax becomes normal to the circle x²+y²-2ax-2by+ c=0 is given by

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$$\begin{gathered} \mathrm{The} \mathrm{information} \mathrm{provided} \mathrm{is} \mathrm{not} \mathrm{very} \mathrm{clear}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{normal} \mathrm{of} \mathrm{circle} \mathrm{pass} \mathrm{through} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \mathrm{We} \mathrm{have}: \\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{ax}-2\mathrm{by}+\mathrm{c}=0 \\ \mathrm{Centre} \mathrm{will} \mathrm{be} \left(-\frac{\left(-2\mathrm{a}\right)}{2},-\frac{\left(-2\mathrm{b}\right)}{2}\right)=\left(\mathrm{a},\mathrm{b}\right) \\ \mathrm{Pair} \mathrm{of} \mathrm{tangents} \mathrm{will} \mathrm{pass} \mathrm{through} \mathrm{centre} \mathrm{of} \mathrm{circle} \mathrm{or} \mathrm{in} \mathrm{other} \mathrm{words}, \mathrm{pair} \mathrm{of} \mathrm{tangents} \\ \mathrm{must} \mathrm{originate} \mathrm{from} \mathrm{centre} \mathrm{of} \mathrm{circle}. \\ \mathrm{We} \mathrm{have} {\mathrm{y}}^2=4\mathrm{ax} \\ \mathrm{S}:{\mathrm{y}}^2-4\mathrm{ax}=0 \\ \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(\mathrm{a},\mathrm{b}\right) \\ {\mathrm{S}}_1: {{\mathrm{y}}_1}^2-4{\mathrm{ax}}_1=0 \\ \mathrm{T}: {\mathrm{yy}}_1-4\mathrm{a}\left(\frac{\mathrm{x}+{\mathrm{x}}_1}{2}\right)=0 \\ \Rightarrow \mathrm{T}: {\mathrm{yy}}_1-2\mathrm{a}\left(\mathrm{x}+{\mathrm{x}}_1\right)=0 \\ \mathrm{Equation} \mathrm{of} \mathrm{pair} \mathrm{of} \mathrm{tangents} \mathrm{is}: \\ {\mathrm{SS}}_1={\mathrm{T}}^2 \\ \Rightarrow \left({\mathrm{y}}^2-4\mathrm{ax}\right)\left({{\mathrm{y}}_1}^2-4{\mathrm{ax}}_1\right)={\left\{{\mathrm{yy}}_1-2\mathrm{a}\left(\mathrm{x}+{\mathrm{x}}_1\right)\right\}}^2 \\ \Rightarrow \left({\mathrm{y}}^2-4\mathrm{ax}\right)\left({\mathrm{b}}^2-4\mathrm{a}.\mathrm{a}\right)={\left\{\mathrm{yb}-2\mathrm{a}\left(\mathrm{x}+\mathrm{a}\right)\right\}}^2 \\ \mathbf{\Rightarrow }\left({\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ax}\right)\left({\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{a}}^{\mathbf{2}}\right)\mathbf{=}{\left\{\mathbf{by}\mathbf{-}\mathbf{2}\mathbf{a}\left(\mathbf{x}\mathbf{+}\mathbf{a}\right)\right\}}^{\mathbf{2}} \end{gathered}$$

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