Question

Find the radius of x^2-6x+8y+y^2=0

Answer 1

Answer:

$\mathrm{Circle\:radius\:given}\:x^2-6x+8y+y^2=0:\quad r=5$

Step-by-step explanation:

$x^2-6x+8y+y^2=0$

$\mathrm{Circle\:Equation}$

• $\left(x-a\right)^2+\left(y-b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)$

$\mathrm{Rewrite}\:x^2-6x+8y+y^2=0\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}$

$x^2-6x+8y+y^2=0$

$\mathrm{Group\:x-variables\:and\:y-variables\:together}$

$\left(x^2-6x\right)+\left(y^2+8y\right)=0$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$

$\left(x^2-6x+9\right)+\left(y^2+8y\right)=9$

$\mathrm{Convert\:to\:square\:form}$

$\left(x-3\right)^2+\left(y^2+8y\right)=9$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$

$\left(x-3\right)^2+\left(y^2+8y+16\right)=9+16$

$\mathrm{Convert\:to\:square\:form}$

$\left(x-3\right)^2+\left(y+4\right)^2=9+16$

$\mathrm{Refine\:}9+16$

$\left(x-3\right)^2+\left(y+4\right)^2=25$

$\mathrm{Rewrite\:in\:standard\:form}$

$\left(x-3\right)^2+\left(y-\left(-4\right)\right)^2=5^2$

$\left(x-3\right)^2+\left(y-\left(-4\right)\right)^2=5^2$

$\mathrm{Therefore\:the\:circle\:properties\:are:}$

$\left(a,\:b\right)=\left(3,\:-4\right),\:r=5$

$\mathrm{And\:the\:radius\:is:}$

$r=5$