Question

$radius\:x^2-6x+8y+y^2=0$

solve

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Answer 1

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

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

$\mathrm{Circle\:Equation}$

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

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

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

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

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

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

$r=5$

Answer 2

Explanation:

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

2

−6x+8y+y

2

=0:r=5

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

2

−6x+8y+y

2

=0

\mathrm{Circle\:Equation}CircleEquation

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

2

+(y−b)

2

=r

2

isthecircleequationwitharadiusr,centeredat(a,b)

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

2

−6x+8y+y

2

=0intheformofthestandardcircleequation

(x-3)^2+(y-(-4))^2=5^2(x−3)

2

+(y−(−4))

2

=5

2

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

(a,\:b)=(3,\:-4),\:r=5(a,b)=(3,−4),r=5

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

r=5r=5