If the line 3x-4y= k cuts the circle x²+y²-4x-8y-5=0 in two points then the limits of k are

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$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ 3\mathrm{x}-4\mathrm{y}=\mathrm{k} \\ \Rightarrow 3\mathrm{x}-4\mathrm{y}-\mathrm{k}=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}-5=0 \\ \mathrm{Center} \mathrm{of} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \mathrm{is} \left(-\mathrm{g},-\mathrm{f}\right) \\ \mathrm{Hence} \mathrm{center} \mathrm{of} \mathrm{given} \mathrm{circle}=\left(2,4\right) \\ \mathrm{Radius}=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \\ =\sqrt{2^2+4^2-\left(-5\right)} \\ =\sqrt{4+16+5} \\ =5 \mathrm{units} \\ \mathrm{If} \mathrm{line} \mathrm{intersects} \mathrm{circle} \mathrm{at} \mathrm{two} \mathrm{points} \mathrm{then} \mathrm{length} \mathrm{of} \mathrm{perpendicular} \mathrm{from} \mathrm{center} \mathrm{of} \mathrm{circle} \mathrm{must} \mathrm{be} \\ \mathrm{less} \mathrm{than} \mathrm{rad}\mathrm{ius} \mathrm{of} \mathrm{circle}. \\ \Rightarrow \frac{\left|3.2-4.4-\mathrm{k}\right|}{\sqrt{3^2+4^2}}<5 \\ \Rightarrow \frac{\left|6-16-\mathrm{k}\right|}{5}<5 \\ \Rightarrow \left|-\left(\mathrm{k}+10\right)\right|<25 \\ \Rightarrow \left|\mathrm{k}+10\right|<25 \\ \mathrm{if} \\ \left|\mathrm{x}\right|<\mathrm{a}, \mathrm{then} \\ -\mathrm{a}<\mathrm{x}<\mathrm{a} \\ \Rightarrow -25<\mathrm{k}+10<25 \\ \Rightarrow -25-10<\mathrm{k}+10-10<25-10 \\ \mathbf{\Rightarrow }\mathbf{-}\mathbf{35}\mathbf{<}\mathbf{k}\mathbf{<}\mathbf{15}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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