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 \( \boxed{2} \).
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.