Determine each of the following for the circle whose equation is \( x^{2}+10 x+y^{2}-12 y+57=0 \). The equation of the circle in standard (center, radius) form is: The center of the circle is: The radius of the circle is:

To determine the center and radius of the circle given by the equation \[ x^{2}+10 x+y^{2}-12 y+57=0, \] 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. ### Step 1: Rearranging the equation We start by moving the constant term to the other side: \[ x^{2} + 10x + y^{2} - 12y = -57. \] ### Step 2: Completing the square Next, we complete the square for the \(x\) and \(y\) terms. **For the \(x\) terms:** 1. Take the coefficient of \(x\), which is \(10\), divide it by \(2\) to get \(5\), and then square it to get \(25\). 2. Add and subtract \(25\): \[ x^{2} + 10x = (x^{2} + 10x + 25) - 25 = (x + 5)^{2} - 25. \] **For the \(y\) terms:** 1. Take the coefficient of \(y\), which is \(-12\), divide it by \(2\) to get \(-6\), and then square it to get \(36\). 2. Add and subtract \(36\): \[ y^{2} - 12y = (y^{2} - 12y + 36) - 36 = (y - 6)^{2} - 36. \] ### Step 3: Substitute back into the equation Now substitute these completed squares back into the equation: \[ ((x + 5)^{2} - 25) + ((y - 6)^{2} - 36) = -57. \] This simplifies to: \[ (x + 5)^{2} + (y - 6)^{2} - 61 = -57. \] Adding \(61\) to both sides gives: \[ (x + 5)^{2} + (y - 6)^{2} = 4. \] ### Step 4: Identify the center and radius Now we can identify the center and radius from the standard form: - The center \((h, k)\) is \((-5, 6)\). - The radius \(r\) is \(\sqrt{4} = 2\). ### Final Answers - The equation of the circle in standard form is: \[ (x + 5)^{2} + (y - 6)^{2} = 4. \] - The center of the circle is: \[ (-5, 6). \] - The radius of the circle is: \[ 2. \]