Answer:
The standard form equation is [tex]\left(x-\left(-4\right)\right)^2+\left(y-\left(-2\right)\right)^2=5^2[/tex].
The center is [tex]\left(h,\:k\right)=\left(-4,\:-2\right)[/tex] and the radius is [tex]r=5[/tex].
Step-by-step explanation:
The standard form of an equation of a circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and the radius is [tex]r[/tex].
To write the equation [tex]x^2+8x+y^2+4y-5=0[/tex] in the form of the standard circle equation you must:
[tex]\mathrm{Move\:the\:five\:to\:the\:right\:side} \\\\x^2+8x+y^2+4y=5\\\\\mathrm{Group\:x-variables\:and\:y-variables\:together}\\\\\left(x^2+8x\right)+\left(y^2+4y\right)=5\\\\\mathrm{Convert}\:x\:\mathrm{to\:square\:form}\\\\\left(x^2+8x+16\right)+\left(y^2+4y\right)=5+16\\\left(x+4\right)^2+\left(y^2+4y\right)=5+16\\\\\mathrm{Convert}\:y\:\mathrm{to\:square\:form}\\\\\left(x+4\right)^2+\left(y^2+4y+4\right)=5+16+4\\\left(x+4\right)^2+\left(y+2\right)^2=5+16+4\\[/tex]
[tex]\left(x+4\right)^2+\left(y+2\right)^2=25\\\\\mathrm{Rewrite\:in\:standard\:form}\\\\\left(x-\left(-4\right)\right)^2+\left(y-\left(-2\right)\right)^2=5^2\\\\\mathrm{Therefore\:the\:circle\:properties\:are:}\\\\\left(h,\:k\right)=\left(-4,\:-2\right),\:r=5[/tex]
