Answer :

Given equation is

[tex]x^2+y^2+8x-2y+15=0[/tex]

Convert this equation into standard form of circle equation.

[tex]\begin{gathered} x^2+2(4)x+4^2+y^2_{}+2(-1)y+1+15-16-1=0 \\ (x+4)^2+(y-1)^2-2=0 \end{gathered}[/tex]

On simplifying,

[tex]\begin{gathered} (x+4)^2+(y-1)^2=2 \\ (x+4)^2+(y-1)^2=(\sqrt[]{2})^2 \end{gathered}[/tex]

Hence the radius of the circle is

[tex]\sqrt[]{2}[/tex]

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