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: \( \boxed{1} \)

Answer 2

To determine the value of \( a \) such that the circles have only one common tangent, we first analyze their equations. The equation of the first circle is \( x^2 + y^2 - 8x - 8y + 28 = 0 \). Rewriting it in standard form, we complete the square:

\( x^2 - 8x + y^2 - 8y + 28 = 0 \)

\( (x^2 - 8x + 16) + (y^2 - 8y + 16) = -28 + 32 \)

\( (x - 4)^2 + (y - 4)^2 = 4 \)

Thus, the center of the first circle is \( (4, 4) \) with radius 2.

The equation of the second circle is \( x^2 + y^2 - 8x - 6y + 25 - a^2 = 0 \). Similarly, we complete the square:

\( (x^2 - 8x + 16) + (y^2 - 6y + 9) = a^2 - 25 + 25 \)

\( (x-4)^2 + (y-3)^2 = a^2 \)

Thus, the center of the second circle is \( (4, 3) \) with radius \( a \).

With the centers \( (4,4) \) and \( (4,3) \), the distance between them is:

\(\sqrt{(4-4)^2 + (4-3)^2} = \sqrt{1} = 1\)

For two circles to have exactly one common tangent, the distance between centers equals the absolute difference of their radii. Since the radius of the first circle is 2, and assuming the radius of the second is \( a \), we have:

\(|2 - a| = 1\)

This gives two scenarios:

\(2 - a = 1 \Rightarrow a = 1\)

\(a - 2 = 1 \Rightarrow a = 3\)

However, for only one tangent, the radii must differ by exactly the center distance; hence the permitted radius based on the context here will be \( a = 1 \).