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

The Correct Option is A

To determine the range of \(r\) for which the circles intersect at precisely two points, we will examine the intersection criteria. The initial circle is defined by the equation \((x + 1)^2 + (y + 2)^2 = r^2\), possessing a center \(C_1 = (-1, -2)\) and a radius \(r_1 = r\). The second circle's equation can be rearranged to \((x - 2)^2 + (y - 2)^2 = 9\), indicating a center \(C_2 = (2, 2)\) and a radius \(r_2 = 3\). The distance \(d\) between the centers \(C_1\) and \(C_2\) is calculated as follows:
\[ d = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
For two circles to intersect at exactly two points, the condition \(|r_1 - r_2| < d < r_1 + r_2\) must be satisfied. Substituting \(r_1 = r\), \(r_2 = 3\), and \(d = 5\) into this condition yields two inequalities:
The first inequality: \(|r - 3| < 5\)
The second inequality: \(5 < r + 3\)
The combined solution to these inequalities is:
\[ 3 < r < 7 \]