Checkpoint Sketch the graph of $$ (x+2)^{2}+(y-2)^{2}

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Checkpoint Sketch the graph of $$ (x+2)^{2}+(y-2)^{2}<16 $$

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To sketch the graph of the inequality $$ (x+2)^{2}+(y-2)^{2}<16 $$, we can follow these steps:

1. **Identify the type of graph**: The expression $$ (x+2)^{2}+(y-2)^{2}=16 $$ represents a circle. The inequality $$ < 16 $$ indicates that we are interested in the area inside this circle.

2. **Determine the center and radius**:
   - The center of the circle is at the point $$ (-2, 2) $$.
   - The radius $$ r $$ can be found by taking the square root of 16, which gives $$ r = 4 $$.

3. **Draw the circle**:
   - Plot the center at $$ (-2, 2) $$.
   - From the center, measure 4 units in all directions (up, down, left, right) to find the points on the circle:
     - Up: $$ (-2, 6) $$
     - Down: $$ (-2, -2) $$
     - Left: $$ (-6, 2) $$
     - Right: $$ (2, 2) $$

4. **Sketch the circle**:
   - Draw a circle with the center at $$ (-2, 2) $$ and passing through the points identified above.

5. **Shade the interior**:
   - Since the inequality is $$ < 16 $$, shade the area inside the circle to indicate that all points within this region satisfy the inequality.

6. **Boundary**:
   - The boundary of the circle itself (where $$ (x+2)^{2}+(y-2)^{2}=16 $$) is not included in the solution set, so you can represent this with a dashed line.

This process will give you a clear sketch of the graph for the inequality $$ (x+2)^{2}+(y-2)^{2}<16 $$.
To sketch the graph of $$ (x+2)^{2}+(y-2)^{2}<16 $$, draw a circle with center at $$ (-2, 2) $$ and radius 4. Shade the area inside the circle, using a dashed line for the boundary.

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