How do you find the center (h,k) and radius r of the circle with the given equation $2 {(x - 2)}^{2} + 2 {(y + 5)}^{2} = 28$ $2(x-2)^2+2(y+5)^2=28$ ?

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Answer 1 · 2016-01-14T00:31:06

Rearrange slightly into standard form and read off $(h, k) = (2, - 5)$ $(h, k) = (2, -5)$ and $r = \sqrt{14}$ $r = sqrt(14)$

If you divide both sides by $2$ $2$ , tweak the monomial in $y$ $y$ a bit and express the right hand side as a square you get the following:

${(x - 2)}^{2} + {(y - (- 5))}^{2} = {(\sqrt{14})}^{2}$ $(x-2)^2+(y-(-5))^2 = (sqrt(14))^2$

This is in the form:

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

with $(h, k) = (2, - 5)$ $(h, k) = (2, -5)$ and $r = \sqrt{14}$ $r = sqrt(14)$

How do you find the center (h,k) and radius r of the circle with the given equation 2(x−2)2+2(y+5)2=282(x-2)^2+2(y+5)^2=28?