Where is the center of the circle given by the equation ${(x - 3)}^{2} + {(y + 2)}^{2} = 9$ $(x - 3)^2 + (y + 2)^2 = 9$ ?

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Answer 1 · 2018-05-25T07:45:17

the centre is $(3, - 2)$ $" "(3,-2)$

an eqn of a circle is given by

${(x - a)}^{2} + {(y - b)}^{2} = r^{2} - - (1)$ $(x-a)^2+(y-b)^2=r^2--(1)$

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

and the radius is $r$ $r$

for

${(x - 3)}^{2} + {(y + 2)}^{2} = 9$ $(x-3)^2+(y+2)^2=9$

comparing with $(1)$ $(1)$

the centre is $(3, - 2)$ $(3,-2)$

as an extra we see the radius is

$r = \sqrt{9} = 3$ $r=sqrt9=3$

Where is the center of the circle given by the equation (x - 3)^2 + (y + 2)^2 = 9(x−3)2+(y+2)2=9(x - 3)^2 + (y + 2)^2 = 9?