Question

a restaurant makes a map of their city on a coordinate plane where each unit represents 1 mile. the restaurant delivers to the area inside the equation $(x + 6)^2+(y - 4)^2 = 9$. select all of the coordinates that represent locations within the delivery area. $(-6,6)$ $(-5,3)$ $(2,-6)$ $(4,-4)$ $(6,-4)$

Explanation:

Step1: Recall the equation of a circle

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. For the equation \((x + 6)^2+(y - 4)^2=9\), the center is \((-6,4)\) and the radius \(r = 3\) (since \(r^2=9\)).

Step2: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

• For the point \((-6,6)\):

\(d=\sqrt{(-6+6)^2+(6 - 4)^2}=\sqrt{0 + 4}=2\). Since \(2<3\), the point \((-6,6)\) is inside the circle.

• For the point \((-5,3)\):

\(d=\sqrt{(-5 + 6)^2+(3 - 4)^2}=\sqrt{1+1}=\sqrt{2}\approx1.41<3\), so the point \((-5,3)\) is inside the circle.

• For the point \((2,-6)\):

\(d=\sqrt{(2 + 6)^2+(-6 - 4)^2}=\sqrt{64 + 100}=\sqrt{164}\approx12.8>3\), so the point \((2,-6)\) is outside the circle.

• For the point \((4,-4)\):

\(d=\sqrt{(4 + 6)^2+(-4 - 4)^2}=\sqrt{100+64}=\sqrt{164}\approx12.8>3\), so the point \((4,-4)\) is outside the circle.

• For the point \((6,-4)\):

\(d=\sqrt{(6 + 6)^2+(-4 - 4)^2}=\sqrt{144 + 64}=\sqrt{208}\approx14.4>3\), so the point \((6,-4)\) is outside the circle.

Answer:

\(\boxed{(-6,6)}\), \(\boxed{(-5,3)}\)