What is the center of the circle given by the equation ${(x - 3)}^{2} + {(y - 9)}^{2} = 16$ $(x - 3)^2 + (y - 9)^2 = 16$ ?

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Answer 1 · 2018-03-02T18:50:16

$(3, 9)$ $(3,9)$

The standard form of the equation for a circle is given by:

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

Where:

$h$ $bbh$ is the $x$ $bbx$ coordinate of the centre.

$k$ $bbk$ is the $y$ $bby$ coordinate of the centre.

$r$ $bbr$ is the radius.

From given equation we can see that the centre is at:

$(h, k) = (3, 9)$ $(h,k)=(3,9)$

What is the center of the circle given by the equation (x−3)2+(y−9)2=16(x - 3)^2 + (y - 9)^2 = 16?