Answer
${{x}^{2}}+{{y}^{2}}+2x+4y-4168.16=0$
Work Step by Step
The equation of the circle formed by the surface of the Earth is ${{x}^{2}}+{{y}^{2}}+2x+4y-4091=0$
The center of the circular orbit is same as the center of the Earth, that is the centre of ${{x}^{2}}+{{y}^{2}}+2x+4y-4091=0$
The weather satellite circles $0.6$ units above the earth’s surface
Therefore, the radius of orbit of satellite is $r+0.6$, where $r$is the radius of the Earth.
Let us find the center and radius of the Earth by wirting its equation in standard form $(x-h)^2+(y-k)^2=r^2$
by completing the squares:
\begin{align*}
x^2+y^2+2x+4y-4091&=0\\
x^2+y^2+2x+4y&=4091\\
(x^2+2x)+(y^2+4y)&=4091\\
(x^2+2x+1)+(y^2+4y+4)&=4091+1+4\\
(x^2+2x+1)+(y^2+4y+4)&=4096\\
(x+1)^2+(y+2)^2&=64^2\\
\end{align*}
The radius here us $64$ units.
This means that the radius of the weather satellite is $64+0.6=64.6$ units
Therefore, the standard equation for the orbit of the satellite is
$${{(x+1)}^{2}}+{{(y+2)}^{2}}={{(64.6)}^{2}}$$
Write in general form by expanding and putting all terms on the left side of the equation:
$$x^2+2x+1+y^2+4y+4=4173.16\\
x^2+y^2+2x+4y+5--4173.16=0\\
x^2+y^2+2x+4y-4168.16=0\\
$$
