Answer:
[tex](x+11)^2+(y+6)^2=324[/tex]
Step-by-step explanation:
[tex]x^2+12y+22x+y^2-167=0\\\mathrm{Circle\:Equation}\\\left(x-a\right)^2+\left(y-b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)\\\mathrm{Rewrite}\:x^2+12y+22x+y^2-167=0\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}\\x^2+12y+22x+y^2-167=0\\\mathrm{Move\:the\:loose\:number\:to\:the\:right\:side}\\x^2+22x+y^2+12y=167\\Group\:x-variables\:and\:y-variables\:together\\\left(x^2+22x\right)+\left(y^2+12y\right)=167[/tex]
[tex]\mathrm{Convert}\:x\:\mathrm{to\:square\:form}\\\left(x^2+22x+121\right)+\left(y^2+12y\right)=167+12\\Convert\:to\:square\:form\\\left(x+11\right)^2+\left(y^2+12y\right)=167+121\\\mathrm{Convert}\:y\:\mathrm{to\:square\:form}\\\left(x+11\right)^2+\left(y^2+12y+36\right)=167+121+36\\Convert\:to\:square\:form\\\left(x+11\right)^2+\left(y+6\right)^2=167+121+36\\\mathrm{Refine\:}167+121+36\\\left(x+11\right)^2+\left(y+6\right)^2=324\\Rewrite\:in\:standard\:form[/tex]
[tex]\left(x-\left(-11\right)\right)^2+\left(y-\left(-6\right)\right)^2=18^2\\\mathrm{Therefore\:the\:circle\:properties\:are:}\\\left(a,\:b\right)=\left(-11,\:-6\right),\:r=18[/tex]
