Question

If the circles \( x^2 + y^2 - 8x - 8y + 28 = 0 \) and \( x^2 + y^2 - 8x - 6y + 25 - a^2 = 0 \) have only one common tangent, then \( a \) is: 

• A. \( a = 4 \)
• B. \( a = 2 \)
• C. \( a = 1 \)
• D. \( a = 5 \)

Hint:

For two circles to have only one common tangent, the difference of their radii must be equal to the distance between their centers.

Answer 1

The Correct Option is C

Step 1: Identify the Centers and Radii 
The given circles are: \[ x^2 + y^2 - 8x - 8y + 28 = 0. \] \[ x^2 + y^2 - 8x - 6y + 25 - a^2 = 0. \] Rewriting both in standard form: For the first circle: \[ (x - 4)^2 + (y - 4)^2 = 4. \] Thus, the center is \( (4,4) \) and radius \( R_1 = 2 \). For the second circle: \[ (x - 4)^2 + (y - 3)^2 = a^2. \] Thus, the center is \( (4,3) \) and radius \( R_2 = a \). 
Step 2: Condition for One Common Tangent 
The distance between the centers is: \[ d = \sqrt{(4-4)^2 + (4-3)^2} = \sqrt{1} = 1. \] For the circles to have only one common tangent, the condition is: \[ R_1 - R_2 = d. \] Substituting values: \[ 2 - a = 1. \] Solving for \( a \): \[ a = 1. \] 
Final Answer: \( \boxed{1} \)