if the straight lines 2x+y+5=0 and 4x+2y+15=0 are the tangents to a circle then radius of this circle is equal to

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$$\begin{gathered} 2\mathrm{x}+\mathrm{y}+5=0...\left(\mathrm{i}\right) \\ \mathrm{Slope}={\mathrm{m}}_1=-2 \\ 4\mathrm{x}+2\mathrm{y}+15=0 \\ \mathrm{Divide} \mathrm{both} \mathrm{sides} \mathrm{by} 2 \\ \Rightarrow 2\mathrm{x}+\mathrm{y}+\frac{15}{2}=0 \\ \mathrm{Slope}={\mathrm{m}}_2=-2 \\ \mathrm{As} {\mathrm{m}}_1={\mathrm{m}}_2, \mathrm{hence} \mathrm{lines} \mathrm{are} \mathrm{parallel} \mathrm{which} \mathrm{means} \mathrm{that} \mathrm{lines} \mathrm{will} \mathrm{be} \mathrm{parallel} \\ \mathrm{to} \mathrm{each} \mathrm{other}. \\ \mathrm{Distance} \mathrm{between} \mathrm{parallel} \mathrm{lines} \mathrm{ax}+\mathrm{by}+{\mathrm{c}}_1=0 \mathrm{and} \mathrm{ax}+\mathrm{by}+{\mathrm{c}}_2=0 \mathrm{is} \\ \frac{\left|{\mathrm{c}}_1-{\mathrm{c}}_2\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{Hence} \mathrm{diameter} \mathrm{will} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{distance} \mathrm{between} \mathrm{lines}. \\ \mathrm{Distance} \mathrm{between} \mathrm{lines}=\frac{\left|{\displaystyle \frac{15}{2}}-2\right|}{\sqrt{2^2+1^2}}=\frac{5}{2\sqrt{5}}=\frac{\sqrt{5}}{2} \\ \mathrm{Radius} =\frac{\mathrm{diameter}}{2}=\frac{\sqrt{5}}{4} \end{gathered}$$

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