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
\( 3 < r < 7 \)
\( 0 < r < 7 \)
\( 5 < r < 9 \)
\( \frac{1}{2} < r < 7 \)
The Correct Option is A
Solution and Explanation
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:
\[ 3 < r < 7 \]
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