How do you find the center and radius of the circle given ${(x - 3)}^{2} + {(y + 7)}^{2} = 50$ $(x-3)^2+(y+7)^2=50$ ?

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Answer 1 · 2016-12-04T04:01:39

The center is $(3, - 7)$ $(3,-7)$ and the radius is $5 \sqrt{2}$ $5sqrt2$ .

Find the center and radius of ${(x - 3)}^{2} + {(y + 7)}^{2} = 50$ $(x-3)^2 + (y+7)^2=50$

The equation of a circle is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2 +(y-k)^2=r^2$

where $(h, k) =$ $(h,k)=$ the center and $r =$ $r=$ the radius.

$\Rightarrow (h, k) = (3, - 7)$ $=> (h,k)=(3, -7)$ (note that the signs are opposite of the signs inside the parentheses)

$r^{2} = 50$ $r^2=50$

To find r, square root both sides

$\sqrt{r^{2}} = \sqrt{50}$ $sqrt(r^2)=sqrt50$

$r = \sqrt{25 \cdot 2}$ $r=sqrt(25*2)$

$r = 5 \sqrt{2}$ $r=5sqrt2$

How do you find the center and radius of the circle given (x-3)^2+(y+7)^2=50(x−3)2+(y+7)2=50(x-3)^2+(y+7)^2=50?