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(x+4)^(2)+(y+2)^(2)=25
The given equation represents a circle in the
xy-plane. If
(-4,y) is a point on the circle and
y > 0, what is the value of
y ?

$(x+4)^{2}+(y+2)^{2}=25$ $\newline$ The given equation represents a circle in the $x y$ -plane. If $(-4, y)$ is a point on the circle and $y>0$ , what is the value of $y$ ?

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Convert equations of conic sections from general to standard form

Full solution

Q.

(x+4)^{2}+(y+2)^{2}=25

\newline

The given equation represents a circle in the

x y

-plane. If

(-4, y)

is a point on the circle and

y>0

, what is the value of

y

?

Plug $x = -4$ : Plug $x = -4$ into the equation to find $y$ .\(\newline\) $((-4)+4)^2 + (y+2)^2 = 25$ \(\newline\) $(0)^2 + (y+2)^2 = 25$ \(\newline\) $(y+2)^2 = 25$
Take square root: Take the square root of both sides to solve for $y+2$ . $\sqrt{(y+2)^2} = \sqrt{25}$ $y+2 = \pm 5$
Choose positive solution: Since $y > 0$ , we choose the positive solution. $\newline$ $y+2 = 5$
Subtract $2$ : Subtract $2$ from both sides to solve for $y$ . $\newline$ $y = 5 - 2$ $\newline$ $y = 3$

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