Where is the center of the graph of ${(x + 3)}^{2} + {(y - 2)}^{2} = 4$ $(x+3)^2+(y-2)^2=4$ ?

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Answer 1 · 2014-09-11T06:34:28

Standard form:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

where $(h, k)$ $(h,k)$ is the center of the circle and $r$ $r$ is the radius of the circle

The center of the circle is $(- 3, 2)$ $(-3,2)$ . You need to use the opposite signs for the numbers used in place of $(h, k)$ $(h,k)$

Where is the center of the graph of (x+3)^2+(y-2)^2=4(x+3)2+(y−2)2=4(x+3)^2+(y-2)^2=4?