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:
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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.
Given are two circles: 1. \( C_1 \), a circle in the third quadrant with radius 3, centered at \( (3, 3) \). 2. \( C_2 \), a circle with center at \( (1, 3) \), which touches \( C_1 \) externally at \( (\alpha, \beta) \). Step 1: Equation of Circle \( C_1 \) The equation for \( C_1 \), with center \( (3, 3) \) and radius 3, is:\[(x - 3)^2 + (y - 3)^2 = 9\] Step 2: Equation of Circle \( C_2 \) The equation for \( C_2 \), with center \( (1, 3) \) and radius \( r_2 \), is:\[(x - 1)^2 + (y - 3)^2 = r_2^2\]Since the circles touch externally, the distance between their centers equals the sum of their radii:\[\text{Distance between centers} = \sqrt{(3 - 1)^2 + (3 - 3)^2} = 2\] Therefore, 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 \( (\beta - \alpha)^2 \) is given as \( \frac{m}{n} \). Using the aforementioned relationship:\[(\beta - \alpha)^2 = 22\] Thus, \( m + n = 22 \).
