How do you graph $(x - 3)^ 2 + (y + 1) ^2 = 9$ ?

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Answer 1 · 2016-02-13T05:37:42

Draw a circle with $(3, -1)$ as center and $3$ as radius.

Standard equation of a circle with $(a, b)$ as center and $r$ as radius is

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

Hence, in the above, $(3, -1)$ is the center and $3$ is the radius.

Hence to draw the graph, just draw a circle with $(3, -1)$ as center and $3$ as radius.

How do you graph (x - 3)^ 2 + (y + 1) ^2 = 9 (x - 3)^ 2 + (y + 1) ^2 = 9 ?