If the line px+qy =1 touches the circle x^2+y^2=r^2, prove that the point (p,q) lies on the circle x^2+y^2=r^-2.

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{px}+\mathrm{qy}-1=0 \\ \mathrm{Equation} \mathrm{of} \mathrm{circle} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \mathrm{Center} \mathrm{is} \left(0,0\right) \mathrm{aand} \mathrm{radius} \mathrm{is} \mathrm{r} \mathrm{units}. \\ \mathrm{As} \mathrm{line} \mathrm{touches} \mathrm{circle}, \mathrm{hence} \mathrm{length} \mathrm{of} \mathrm{perpendicular} \mathrm{from} \mathrm{centre} \mathrm{of} \mathrm{cirlce} \\ \mathrm{to} \mathrm{the} \mathrm{line} \mathrm{must} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle}. \\ \Rightarrow \frac{\left|\mathrm{p}\times 0+\mathrm{q}\times 0-1\right|}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}=\mathrm{r} \\ \Rightarrow \frac{\left|-1\right|}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}=\mathrm{r} \\ \Rightarrow \frac{1}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}=\mathrm{r} \\ \mathrm{Squarring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}: \\ \Rightarrow \frac{1}{{\mathrm{p}}^2+{\mathrm{q}}^2}={\mathrm{r}}^2 \\ \Rightarrow {\mathrm{p}}^2+{\mathrm{q}}^2=\frac{1}{{\mathrm{r}}^2} \\ \Rightarrow {\mathrm{p}}^2+{\mathrm{q}}^2={\mathrm{r}}^{-2} \\ \mathrm{If} \mathrm{we} \mathrm{replace} \mathrm{p} \mathrm{with} \mathrm{x} \mathrm{and} \mathrm{q} \mathrm{with} \mathrm{y}, \mathrm{then}: \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^{-2} \\ \mathrm{Hence} \mathrm{point} \left(\mathrm{p},\mathrm{q}\right) \mathrm{lies} \mathrm{on} \mathrm{cirlce} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^{-2} \end{gathered}$$
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