How do you graph ${(x + 5)}^{2} + {(y - 2)}^{2} = 9$ $(x+5)^2+(y-2)^2=9$ ?

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Answer 1 · 2017-02-12T08:46:35

This is a circle with centre $(- 5, 2)$ $(-5, 2)$ and radius $3$ $3$

Given:

${(x + 5)}^{2} + {(y - 2)}^{2} = 9$ $(x+5)^2+(y-2)^2=9$

This can be rewritten slightly as:

${(x - (- 5))}^{2} + {(y - 2)}^{2} = 3^{2}$ $(x-(-5))^2+(y-2)^2 = 3^2$

which is in the standard form:

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

which is the equation of a circle with centre $(h, k) = (- 5, 2)$ $(h, k) = (-5, 2)$ and radius $r = 3$ $r=3$ .

graph{((x+5)^2+(y-2)^2-9)((x+5)^2+(y-2)^2-0.02) = 0 [-15.75, 4.25, -3.12, 6.88]}

How do you graph (x+5)^2+(y-2)^2=9(x+5)2+(y−2)2=9(x+5)^2+(y-2)^2=9?