Question

The shortest distance between the circles \( (x-1)^2 + (y+2)^2 = 1 \) and \( (x+2)^2 + (y-2)^2 = 4 \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)
• E. \( 5 \)

Hint:

If \( |C_1C_2| < |r_1 - r_2| \), one circle is inside the other and the shortest distance is \( |r_1 - r_2| - |C_1C_2| \).

Answer 1

The Correct Option is B

Concept: The shortest distance between two circles with centers \( C_1, C_2 \) and radii \( r_1, r_2 \) is: \[ d_{min} = |C_1C_2| - (r_1 + r_2) \] provided the circles are external to each other.

Step 1: Identify centers and radii.
Circle 1: \( C_1 = (1, -2) \), \( r_1 = \sqrt{1} = 1 \). Circle 2: \( C_2 = (-2, 2) \), \( r_2 = \sqrt{4} = 2 \).

Step 2: Calculate the distance between centers.
Using distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \): \[ |C_1C_2| = \sqrt{(-2-1)^2 + (2 - (-2))^2} \] \[ = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = 5 \]

Step 3: Find the shortest distance.
Sum of radii \( r_1 + r_2 = 1 + 2 = 3 \). Since \( |C_1C_2| > r_1 + r_2 \) (\( 5 > 3 \)), the circles are separate. \[ d_{min} = 5 - 3 = 2 \]