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 \( \boxed{2} \).
What is the diameter of the circle in the figure ? 
Consider the above figure and read the following statements.
Statement 1: The length of the tangent drawn from the point P to the circle is 24 centimetres. If OP is 25 centimetres, then the radius of the circle is 7 centimetres.
Statement 2: A tangent to a circle is perpendicular to the radius through the point of contact.
Now choose the correct answer from those given below. 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]