How do you find the center and radius of the circle given ${(x - 3)}^{2} + {(y - 1)}^{2} = 25$ $(x-3)^2+(y-1)^2=25$ ?

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Answer 1 · 2016-11-08T21:57:20

centre is $(3, 1)$ $(3,1)$

$r = 5$ $r=5$

the standard equation of a circle is

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

where $(a, b)$ $(a, b)$ is the centre of the circle

and $r$ $r$ its radius.

${(x - 3)}^{2} + {(y - 1)}^{2} = 25$ $(x-3)^2+(y-1)^2=25$

its centre is $(3, 1)$ $(3,1)$

$r = \sqrt{25} = 5$ $r=sqrt25=5$

How do you find the center and radius of the circle given (x−3)2+(y−1)2=25(x-3)^2+(y-1)^2=25?