Graph the circle. \[ x^{2}+y^{2}+4 x-8 y+4=0 \]

Identify the conic by following steps: - step0: Find the standard equation of the circle: \(x^{2}+y^{2}+4x-8y+4=0\) - step1: Move the constant to the right side: \(x^{2}+y^{2}+4x-8y=0-4\) - step2: Add or subtract the terms: \(x^{2}+y^{2}+4x-8y=-4\) - step3: Rearrange the terms: \(x^{2}+4x+y^{2}-8y=-4\) - step4: Add the same value to both sides: \(x^{2}+4x+4+y^{2}-8y=-4+4\) - step5: Factor the expression: \(\left(x+2\right)^{2}+y^{2}-8y=-4+4\) - step6: Add the numbers: \(\left(x+2\right)^{2}+y^{2}-8y=0\) - step7: Add the same value to both sides: \(\left(x+2\right)^{2}+y^{2}-8y+16=16\) - step8: Rewrite the expression: \(\left(x+2\right)^{2}+\left(y-4\right)^{2}=16\) To graph the circle given by the equation \[ x^{2}+y^{2}+4 x-8 y+4=0, \] we first rewrite it in standard form. 1. **Rearranging the equation:** We can rearrange the equation to isolate the terms involving \(x\) and \(y\): \[ x^{2} + 4x + y^{2} - 8y + 4 = 0. \] 2. **Completing the square:** - For the \(x\) terms: \(x^{2} + 4x\) can be rewritten as \((x + 2)^{2} - 4\). - For the \(y\) terms: \(y^{2} - 8y\) can be rewritten as \((y - 4)^{2} - 16\). Substituting these back into the equation gives: \[ (x + 2)^{2} - 4 + (y - 4)^{2} - 16 + 4 = 0. \] Simplifying this, we have: \[ (x + 2)^{2} + (y - 4)^{2} - 16 = 0, \] which leads to: \[ (x + 2)^{2} + (y - 4)^{2} = 16. \] 3. **Identifying the center and radius:** The standard form of a circle is \((x - h)^{2} + (y - k)^{2} = r^{2}\), where \((h, k)\) is the center and \(r\) is the radius. From our equation: - Center: \((-2, 4)\) - Radius: \(\sqrt{16} = 4\) 4. **Graphing the circle:** The circle is centered at \((-2, 4)\) with a radius of \(4\). To visualize this, you would plot the center at \((-2, 4)\) and draw a circle that extends \(4\) units in all directions from this point. Unfortunately, I cannot create a plot directly, but you can use graphing software or graphing calculators to visualize the circle based on the center and radius provided.