Which of these points are 5 units away from $$ (-6,-3) $$ ? Select all that appl \begin{tabular}{lll|l|}$$ (-2,0) $$ & $$ (-3,-7) $$ & $$ (-2,-7) $$\end{tabular}

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We will use the distance formula to find the distance between a given point $$(x, y)$$ and the center $$(-6, -3)$$. The distance formula is

$$
d = \sqrt{(x - (-6))^2 + (y - (-3))^2} = \sqrt{(x + 6)^2 + (y + 3)^2}.
$$

We want the distance $$ d $$ to equal 5. Thus, we need

$$
\sqrt{(x + 6)^2 + (y + 3)^2} = 5.
$$

Squaring both sides gives

$$
(x + 6)^2 + (y + 3)^2 = 25.
$$

Now, we will check each given point.

1. For $$(-2, 0)$$:
   - Compute $$x + 6$$: $$-2 + 6 = 4$$.
   - Compute $$y + 3$$: $$0 + 3 = 3$$.
   - Then,
     
     $$
     (4)^2 + (3)^2 = 16 + 9 = 25.
     $$
     
   - Since $$25 = 25$$, the point $$(-2,0)$$ is $$5$$ units away.

2. For $$(-3,-7)$$:
   - Compute $$x + 6$$: $$-3 + 6 = 3$$.
   - Compute $$y + 3$$: $$-7 + 3 = -4$$.
   - Then,
     
     $$
     (3)^2 + (-4)^2 = 9 + 16 = 25.
     $$
     
   - Since $$25 = 25$$, the point $$(-3,-7)$$ is $$5$$ units away.

3. For $$(-2,-7)$$:
   - Compute $$x + 6$$: $$-2 + 6 = 4$$.
   - Compute $$y + 3$$: $$-7 + 3 = -4$$.
   - Then,
     
     $$
     (4)^2 + (-4)^2 = 16 + 16 = 32.
     $$
     
   - Since $$32 \neq 25$$, the point $$(-2,-7)$$ is not $$5$$ units away.

Thus, the points that are $$5$$ units away from $$(-6,-3)$$ are $$(-2,0)$$ and $$(-3,-7)$$.
The points $$(-2,0)$$ and $$(-3,-7)$$ are 5 units away from $$(-6,-3)$$.

Which of these points are 5 units away from ? Select all that appl

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