Answer:
[tex]\mathrm{Circle\:with\:center\:at}\:\left(13,\:1\right)\:\mathrm{and\:radius}\:r=2[/tex]
Step-by-step explanation:
[tex]\left(x-13\right)^2+\left(y-1\right)^2=4\\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}\:\left(x-13\right)^2+\left(y-1\right)^2=4\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}\\\left(x-13\right)^2+\left(y-1\right)^2=2^2\\Therefore\:the\:circle\:properties\:are:\\\left(a,\:b\right)=\left(13,\:1\right),\:r=2[/tex]
