slopes of tangents through (7,1) to the circle x^2+y^2=25 satisfy the equation

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$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2=25 \\ \Rightarrow {\left(\mathrm{x}-0\right)}^2+{\left(\mathrm{y}-0\right)}^2=5^2 \\ \mathrm{Hence} \mathrm{centre} \mathrm{is} \left(0,0\right) \mathrm{and} \mathrm{radius} \mathrm{is} \mathrm{a}=5 \mathrm{units} \\ \mathrm{Any} \mathrm{tangent} \mathrm{of} \mathrm{m} \mathrm{slope} \mathrm{to} \mathrm{above} \mathrm{cirlce} \mathrm{is} \\ \mathrm{y}=\mathrm{mx}\pm \mathrm{a}\sqrt{1+{\mathrm{m}}^2} \\ \mathrm{a}=5 \\ \Rightarrow \mathrm{y}=\mathrm{mx}\pm 5\sqrt{1+{\mathrm{m}}^2} \\ \mathrm{Tangent} \mathrm{passes} \mathrm{through} \left(7,1\right) \\ \Rightarrow 1=7\mathrm{m}\pm 5\sqrt{1+{\mathrm{m}}^2} \\ \Rightarrow 1-7\mathrm{m}=\pm 5\sqrt{1+{\mathrm{m}}^2} \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\left(1-7\mathrm{m}\right)}^2=25\left(1+{\mathrm{m}}^2\right) \\ \Rightarrow 1+49{\mathrm{m}}^2-14\mathrm{m}=25+25{\mathrm{m}}^2 \\ \Rightarrow 24{\mathrm{m}}^2-14\mathrm{m}-24=0 \\ \Rightarrow 12{\mathrm{m}}^2-7\mathrm{m}-12=0 \end{gathered}$$

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