First, rewrite the equation of the circle in standard form by completing the square for both \( x \) and \( y \).
The equation is: \[ 4x^2 + 4y^2 - 12x + 8y = 0. \] Divide through by 4: \[ x^2 + y^2 - 3x + 2y = 0. \]
Now complete the square for \( x \) and \( y \): \[ x^2 - 3x + \left(\frac{3}{2}\right)^2 + y^2 + 2y + 1 = \left(\frac{3}{2}\right)^2 + 1. \]
Simplifying: \[ \left(x - \frac{3}{2}\right)^2 + (y + 1)^2 = \frac{9}{4} + 1 = \frac{13}{4}. \]
Thus, the equation of the circle is: \[ \left(x - \frac{3}{2}\right)^2 + (y + 1)^2 = \frac{13}{4}. \]
The radius \( r \) is the square root of \( \frac{13}{4} \), which is \( \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \). Thus, the radius is \( \boxed{2} \).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: