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 \( \boxed{2} \).

In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity: 
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 