Chapter 1.5, Problem 53AYU

To determine

To find: The equation of the weather satellite that circles 0.6 unit above earth with the centre of its circular orbit at the centre of earth.

Answer to Problem 53AYU

Solution:

$x²+2x+y²+4y -4168.16=0$

Explanation of Solution

Given:

Equation of the circle $x²+y²+2x+4y-4091=0$

Equation of the circle $x²+y²+2x+4y-4091=0$

$\left(x²+2x\right)+\left(y²+4y\right)-4091=0$

$\left(x²+2x\right)+\left(y²+4y\right)-4091+4091=0+4091$   Add 4091 to both sides.

$\left(x²+2x\right)+\left(y²+4y\right)=4091$

$\left(x²+2x+1\right)+\left(y²+4y+4\right)=4091+5$   Completion of squares

$\left(x+1\right)²+\left(y+2\right)²=4096$

$\left(x+1\right)²+\left(y+2\right)²=\left(64\right)²$

Compare this equation with the following standard equation of the circle to find the centre and radius of earth:

$\left(x-h\right)²+\left(y-k\right)²=\left(r\right)²$

Therefore, the centre of the earth is $\left(-1,-2\right)$ and radius is 64 units.

Now the centre of the orbit of the satellite is $\left(-1,-2\right)$ and radius is 64.6 units. The equation of the orbit of the satellite is:

$\left(x+1\right)²+\left(y+2\right)²=\left(64.6\right)²$

$\left(x+1\right)²+\left(y+2\right)²=4173.16$

$x²+2x+1+y²+4y+4 =4173.16$

$x²+2x+y²+4y+5=4173.16$

$x²+2x+y²+4y+5-5=4173.16-5$   Subtract 5 from both sides

$x²+2x+y²+4y =4168.16$

$x²+2x+y²+4y -4168.16 =4168.16-4168.16$   Subtract 4168.16 from both sides

$x²+2x+y²+4y-4168.16 =0$