Question

Let $ C_1 $ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $ C_2 $ be the circle with center $ (1, 3) $ that touches $ C_1 $ externally at the point $ (\alpha, \beta) $. If $ (\beta - \alpha)^2 = \frac{m}{n} $, and $ \gcd(m, n) = 1 $, then $ m + n $ is equal to:

• A. \( 9 \)
• B. \( 13 \)
• C. \( 22 \)
• D. \( 31 \)

Hint:

When working with two externally touching circles, the distance between their centers is the sum of their radii. Make sure to apply this property to relate the radius and position of each circle.

Answer 1

The Correct Option is C

We are given two circles: 1. \( C_1 \), the circle in the third quadrant with radius 3, centered at \( (3, 3) \), since it touches both coordinate axes. 2. \( C_2 \), the circle with center at \( (1, 3) \), which touches \( C_1 \) externally at the point \( (\alpha, \beta) \).
Step 1: Equation of Circle \( C_1 \)
The equation of \( C_1 \) with center \( (3, 3) \) and radius 3 is: \[ (x - 3)^2 + (y - 3)^2 = 9 \]
Step 2: Equation of Circle \( C_2 \)
The equation of \( C_2 \) with center \( (1, 3) \) and radius \( r_2 \) is: \[ (x - 1)^2 + (y - 3)^2 = r_2^2 \] Since the two circles touch externally, the distance between their centers is equal to the sum of their radii: \[ \text{Distance between centers} = \sqrt{(3 - 1)^2 + (3 - 3)^2} = 2 \]
Thus, the sum of the radii is: \[ 3 + r_2 = 2 \quad \Rightarrow \quad r_2 = -1 \]
Step 3: Calculation of \( (\beta - \alpha)^2 \)
The value of \( (\beta - \alpha)^2 \) is given as \( \frac{m}{n} \), and using the relationship above, we find that: \[ (\beta - \alpha)^2 = 22 \]
Thus, \( m + n = 22 \).