A circle has the equation ${(x + 1)}^{2} + {(y - 3)}^{2} = 16$ $(x+1)^2+(y-3)^2=16$ . What are the coordinates of its center and length of its radius?

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Answer 1 · 2016-02-13T21:45:13

The center is $C = (- 1; 3)$ $C=(-1;3)$ the radius is $r = 4$ $r=4$

The equation is in form of ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$ , so the coordinates and the radius are:

$C = (a, b)$ $C=(a,b)$ , radius $r$ $r$

A circle has the equation (x+1)^2+(y-3)^2=16(x+1)2+(y−3)2=16(x+1)^2+(y-3)^2=16. What are the coordinates of its center and length of its radius?