How do you find the center and the length of the radius of $x^{2} + y^{2} = 121$ $x^2 + y^2 = 121$ ?

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Answer 1 · 2018-05-13T19:27:44

$centre (0, 0)$ $"centre "(0,0)$

$r = 11$ $r=11$

the standard equation of a circle centre $(a, b)$ $( a,b)$ and radius $r$ $r$

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

for

$x^{2} + y^{2} = 121$ $x^2+y^2=121$

the centre is $(0, 0)$ $(0,0)$

radius $r = \sqrt{121} = 11$ $r=sqrt121=11$

How do you find the center and the length of the radius of x^2 + y^2 = 121x2+y2=121x^2 + y^2 = 121?