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Answer :
The equation
[tex]$$
(x+10)^2+(y-8)^2=16
$$[/tex]
is in the form
[tex]$$
(x-h)^2+(y-k)^2=r^2,
$$[/tex]
where the center of the circle is [tex]$(h,k)$[/tex] and the radius is [tex]$r$[/tex]. In our equation, comparing we have
[tex]$$
h=-10,\quad k=8,\quad \text{and}\quad r^2=16.
$$[/tex]
Taking the square root of [tex]$r^2$[/tex] gives
[tex]$$
r=\sqrt{16}=4.
$$[/tex]
Since the parking lot light is located at the center of the circle, the greatest distance at which motion is detected is exactly the radius of the circle. Therefore, the greatest distance a person can be from the parking lot light and still be detected is [tex]$4$[/tex] feet.
[tex]$$
(x+10)^2+(y-8)^2=16
$$[/tex]
is in the form
[tex]$$
(x-h)^2+(y-k)^2=r^2,
$$[/tex]
where the center of the circle is [tex]$(h,k)$[/tex] and the radius is [tex]$r$[/tex]. In our equation, comparing we have
[tex]$$
h=-10,\quad k=8,\quad \text{and}\quad r^2=16.
$$[/tex]
Taking the square root of [tex]$r^2$[/tex] gives
[tex]$$
r=\sqrt{16}=4.
$$[/tex]
Since the parking lot light is located at the center of the circle, the greatest distance at which motion is detected is exactly the radius of the circle. Therefore, the greatest distance a person can be from the parking lot light and still be detected is [tex]$4$[/tex] feet.
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