What is the center, radius and diameter of this circle equation? x^2+ (y-6)^2 =169

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Answer 1 · 2018-05-09T18:24:55

$C (0, 6), R = 13, D = 26$ $C(0, 6), R=13, D=26$

$x^{2} + {(y - 6)}^{2} = 169$ $x^2+(y-6)^2=169$

The standard equation of the circle center at: $(h, k)$ $(h, k)$ and Radius $R$ $R$ :

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

Thus in this case:

${(x - 0)}^{2} + {(y - 6)}^{2} = {(13)}^{2}$ $(x-0)^2+(y-6)^2=(13)^2$

Center at $(0, 6)$ $(0, 6)$ , Radius is: $13$ $13$ and the Diameter is just:

$D = 2 \cdot R = 2 \cdot 13 = 26$ $D=2*R=2*13=26$