Give the center and radius of the circle represented by the equation. \[ x^{2}+y^{2}+8 x+6 y+16=0 \] 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 represented by the equation \[ x^{2}+y^{2}+8 x+6 y+16=0, \] we first need to rewrite the equation in standard form. The standard form of a circle's equation is \[ (x - h)^2 + (y - k)^2 = r^2, \] where \((h, k)\) is the center and \(r\) is the radius. 1. **Rearranging the equation**: Start by moving the constant term to the other side: \[ x^{2} + y^{2} + 8x + 6y = -16. \] 2. **Completing the square**: We will complete the square for the \(x\) and \(y\) terms. - For the \(x\) terms: \(x^{2} + 8x\) - Take half of the coefficient of \(x\) (which is 8), square it: \((\frac{8}{2})^2 = 16\). - Thus, \(x^{2} + 8x\) can be rewritten as \((x + 4)^{2} - 16\). - For the \(y\) terms: \(y^{2} + 6y\) - Take half of the coefficient of \(y\) (which is 6), square it: \((\frac{6}{2})^2 = 9\). - Thus, \(y^{2} + 6y\) can be rewritten as \((y + 3)^{2} - 9\). 3. **Substituting back into the equation**: \[ (x + 4)^{2} - 16 + (y + 3)^{2} - 9 = -16. \] Simplifying this gives: \[ (x + 4)^{2} + (y + 3)^{2} - 25 = -16. \] Adding 25 to both sides: \[ (x + 4)^{2} + (y + 3)^{2} = 9. \] 4. **Identifying the center and radius**: From the equation \((x + 4)^{2} + (y + 3)^{2} = 9\), we can see that: - The center \((h, k)\) is \((-4, -3)\). - The radius \(r\) is \(\sqrt{9} = 3\). Thus, the center of the circle is \[ \boxed{(-4, -3)} \] and the radius of the circle is \[ \boxed{3}. \]