[tex]radius\:x^2-6x+8y+y^2=0[/tex]
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[tex]radius\:x^2-6x+8y+y^2=0[/tex]
solve
want return gift
return gift = follow @RockstarPratheek
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[tex]\mathrm{Circle\:radius\:given}\:x^2-6x+8y+y^2=0:\quad r=5[/tex]
[tex]x^2-6x+8y+y^2=0[/tex]
[tex]\mathrm{Circle\:Equation}[/tex]
[tex]\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)[/tex]
[tex]\mathrm{Rewrite}\:x^2-6x+8y+y^2=0\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}[/tex]
[tex]\left(x-3\right)^2+\left(y-\left(-4\right)\right)^2=5^2[/tex]
[tex]\mathrm{Therefore\:the\:circle\:properties\:are:}[/tex]
[tex]\left(a,\:b\right)=\left(3,\:-4\right),\:r=5[/tex]
[tex]\mathrm{And\:the\:radius\:is:}[/tex]
[tex]r=5[/tex]
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