Key Concepts and Formulas

• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Condition for Two Circles to Intersect at Two Distinct Points: Two circles with radii $r_1$ and $r_2$ and distance between their centers $d$ intersect at two distinct points if and only if $|r_1 - r_2| < d < r_1 + r_2$.

Step-by-Step Solution

Step 1: Identify the centers and radii of the given circles.

The first circle is given by $(x+1)^2 + (y+2)^2 = r^2$. Comparing this to the standard form $(x-h)^2 + (y-k)^2 = R^2$, we see that the center is $C_1 = (-1, -2)$ and the radius is $r_1 = r$.

The second circle is given by $x^2 + y^2 - 4x - 4y + 4 = 0$. We complete the square to rewrite this in standard form: $(x^2 - 4x) + (y^2 - 4y) + 4 = 0$ $(x^2 - 4x + 4) + (y^2 - 4y + 4) + 4 - 4 - 4 = 0$ $(x-2)^2 + (y-2)^2 = 4 = 2^2$ Thus, the center is $C_2 = (2, 2)$ and the radius is $r_2 = 2$.

Step 2: Calculate the distance between the centers of the two circles.

The distance $d$ between the centers $C_1 = (-1, -2)$ and $C_2 = (2, 2)$ is given by the distance formula: $d = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{(2+1)^2 + (2+2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 3: Apply the condition for two circles to intersect at two distinct points.

For the two circles to intersect at exactly two distinct points, we must have $|r_1 - r_2| < d < r_1 + r_2$ Substituting $r_1 = r$, $r_2 = 2$, and $d = 5$, we get $|r - 2| < 5 < r + 2$ This inequality can be split into two inequalities: $|r - 2| < 5 \quad \text{and} \quad 5 < r + 2$

Step 4: Solve the first inequality, $|r - 2| < 5$.

The inequality $|r - 2| < 5$ is equivalent to $-5 < r - 2 < 5$ Adding 2 to all parts of the inequality, we get $-5 + 2 < r < 5 + 2$ $-3 < r < 7$ Since $r$ is a radius, it must be positive, so we have $0 < r < 7$.

Step 5: Solve the second inequality, $5 < r + 2$.

Subtracting 2 from both sides, we get $5 - 2 < r$ $3 < r$

Step 6: Combine the results from Steps 4 and 5.

We have $0 < r < 7$ and $3 < r$. Combining these inequalities, we get $3 < r < 7$

Step 7: Check the options given.

The options are: (A) $\frac{1}{2} < r < 7$ (B) $3 < r < 7$ (C) $5 < r < 9$ (D) $0 < r < 7$

The inequality $3 < r < 7$ matches option (B).

Common Mistakes & Tips

• Remember that the radius of a circle must be positive, so $r > 0$. This helps to eliminate negative values when solving inequalities involving $r$.
• When dealing with absolute value inequalities like $|x| < a$, remember to split it into two inequalities: $-a < x < a$.
• Be careful when completing the square to find the center and radius of a circle. Double-check your calculations to avoid errors.

Summary

We found the centers and radii of the two given circles. Then, we used the condition for two circles to intersect at two distinct points, $|r_1 - r_2| < d < r_1 + r_2$, where $r_1$ and $r_2$ are the radii and $d$ is the distance between the centers. Substituting the known values and solving the resulting inequalities, we found that $3 < r < 7$. This corresponds to option (B).

The final answer is \boxed{3 < r < 7}, which corresponds to option (B).