Thread: Conics-Circle Question

To organize a military exercise, a drill sergeant designates an enemy territory represented by the inequality (x-2)^2+(y+5)^2 (less than or equal to 9) on the map of the base, whose axes are scaled in kilometers. If the soldiers respect a no man's land of 0.1km, what relation represents the friendly territory?

Plz explain.
THX in advance

Originally Posted by Solid8Snake

To organize a military exercise, a drill sergeant designates an enemy territory represented by the inequality (x-2)^2+(y+5)^2 (less than or equal to 9) on the map of the base, whose axes are scaled in kilometers. If the soldiers respect a no man's land of 0.1km, what relation represents the friendly territory?

Plz explain.
THX in advance

The equation of a cirlce is

$\displaystyle (x-2)^2+(y+5)^2\le 3^2 $ where 3 is the radius of the circle

If they respect no mans land it will incerase by .1 so the new equaition would be

$\displaystyle (x-2)^2+(y+5)^2 \le (3.1)^2 $

well according to the answer booklet its 9.61. but why is that?

Originally Posted by Solid8Snake

well according to the answer booklet its 9.61. but why is that?

look at my above post again!

$\displaystyle (3.1)^2=....$

thx a million, sorry about my other post im a bit tired

Originally Posted by Solid8Snake

To organize a military exercise, a drill sergeant designates an enemy territory represented by the inequality (x-2)^2+(y+5)^2 (less than or equal to 9) on the map of the base, whose axes are scaled in kilometers. If the soldiers respect a no man's land of 0.1km, what relation represents the friendly territory?

Plz explain.
THX in advance

Your inequality describes the area of a circle (circle line included) with the center C(2, -5) and the radius r = 3.

If the soldiers keep a distance of more than 3.1 km to C then they are in the friendly area:

$\displaystyle (x-2)^2+(y+5)^2 > (3.1)^2$