Question

a large garden has a set of underground sprinklers. suppose the center of the garden is located at the origin of a grid with intervals measured in feet. one sprinkler is positioned at (2, 4) and reaches a radius of 8 feet. the second sprinkler is positioned at (-6, -4) and reaches a radius of 6 feet. the system of inequalities represents this scenario.
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$$\begin{cases}(x - 2)^2+(y - 4)^2\\leq64\\(x + 6)^2+(y + 4)^2\\leq36\\end{cases}$$

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which location is in the region for plants receiving the maximum amount of water?
(-3, -1)
(-7, 1)
(1, -5)
(2, 1)

Explanation:

Step1: Recall distance - formula for a point from the center of a circle

The equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). For a point \((x,y)\) to be inside or on the circle, \((x - h)^2+(y - k)^2\leq r^2\). We need to check which point satisfies both inequalities \((x - 2)^2+(y - 4)^2\leq64\) and \((x + 6)^2+(y + 4)^2\leq36\).

Step2: Check point \((-3,-1)\)

For the first circle with center \((2,4)\) and \(r = 8\): \((-3-2)^2+(-1 - 4)^2=(-5)^2+(-5)^2=25 + 25=50\leq64\). For the second circle with center \((-6,-4)\) and \(r = 6\): \((-3+6)^2+(-1 + 4)^2=3^2+3^2=9 + 9=18\leq36\).

Step3: Check point \((-7,1)\)

For the first circle: \((-7-2)^2+(1 - 4)^2=(-9)^2+(-3)^2=81+9 = 90>64\), so it is not in the region.

Step4: Check point \((1,-5)\)

For the second circle: \((1 + 6)^2+(-5 + 4)^2=7^2+(-1)^2=49 + 1=50>36\), so it is not in the region.

Step5: Check point \((2,1)\)

For the second circle: \((2+6)^2+(1 + 4)^2=8^2+5^2=64 + 25=89>36\), so it is not in the region.

Answer:

\((-3,-1)\)