Give the center and radius of the circle represented by the equation. \[ 4 x^{2}+4 x+4 y^{2}+24 y-363=0 \] The center of the circle is \( \square \). (Type an ordered pair, using integers or fractions.) The radius of the circle is \( \square \). (Type an integer or a simplified fraction.)

To find the center and radius of the circle given by the equation \( 4x^2 + 4x + 4y^2 + 24y - 363 = 0 \), we need to complete the square for both \( x \) and \( y \) terms. The standard form of the equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Let's start by grouping the \(x\) and \(y\) terms and then completing the square for each group: \[ 4x^2 + 4x + 4y^2 + 24y = 363 \] First, we divide the entire equation by 4 to simplify it: \[ x^2 + x + y^2 + 6y = \frac{363}{4} \] Now, we complete the square for \(x\) and \(y\): For \(x\): \[ x^2 + x = (x + \frac{1}{2})^2 - (\frac{1}{2})^2 \] For \(y\): \[ y^2 + 6y = (y + 3)^2 - 3^2 \] Substitute these expressions back into the equation: \[ (x + \frac{1}{2})^2 - \frac{1}{4} + (y + 3)^2 - 9 = \frac{363}{4} \] Now, we combine the constants on the left side: \[ (x + \frac{1}{2})^2 + (y + 3)^2 - \frac{1}{4} - 9 = \frac{363}{4} \] \[ (x + \frac{1}{2})^2 + (y + 3)^2 - \frac{37}{4} = \frac{363}{4} \] Add \(\frac{37}{4}\) to both sides to isolate the squared terms: \[ (x + \frac{1}{2})^2 + (y + 3)^2 = \frac{363}{4} + \frac{37}{4} \] \[ (x + \frac{1}{2})^2 + (y + 3)^2 = \frac{400}{4} \] \[ (x + \frac{1}{2})^2 + (y + 3)^2 = 100 \] Now we have the equation in the standard form of a circle: \[ (x - (-\frac{1}{2}))^2 + (y - (-3))^2 = 10^2 \] From this, we can see that the center of the circle is \((-1/2, -3)\) and the radius is \(10\). The center of the circle is \( \boxed{(-\frac{1}{2}, -3)} \). The radius of the circle is \( \boxed{10} \).