Question

If the circles $(x+1)^2 + (y+2)^2 = r^2$ and $x^2 + y^2 - 4x - 4y + 4 = 0$ intersect at exactly two distinct points, then

• A. $3 < r < 7$
• B. $0 < r < 7$
• C. $5 < r < 9$
• D. $\frac{1}{2} < r < 7$

Answer 1

Answer: (A)

$3 < r < 7$

Answer 2

Solution: To find the range of r for which the circles intersect at exactly two points, we analyze the conditions for intersection.
The first circle has equation $(x + 1)^2 + (y + 2)^2 = r^2$, with center $C_1 = (-1, -2)$ and radius $r_1 = r$.
The second circle can be rewritten as $(x - 2)^2 + (y - 2)^2 = 9$, with center $C_2 = (2, 2)$ and radius $r_2 = 3$.
The distance $d$ between $C_1$ and $C_2$ is:
$d = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
For two circles to intersect at exactly two points, the condition $|r_1 - r_2| < d < r_1 + r_2$ must hold. Substitute $r_1 = r$, $r_2 = 3$, and $d = 5$:
First inequality: $|r - 3| < 5$
Second inequality: $5 < r + 3$
Combining these results, we get:
$3 < r < 7$