Question

What’s the correct answer for this?

Answer 1

`(x+11)^2+(y+6)^2=324`

Explanation:

`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`

`\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`

`\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`

Answer 2

Second option is the correct answer

`(x+11)^2+(y+6)^2 = 324`

Explanation:

`x^(2) +12y+22x+y^2-167=0\\x^(2)+22x +y^2 +12y-167=0\\(x^(2)+22x +121)-121 +(y^2 +12y+36)-36-167=0\\(x+11)^2+(y+6)^2-324 = 0\\\huge\red{\boxed{(x+11)^2+(y+6)^2 = 324}}\\`