Question

exercises 111–122: (see examples 12 and 13.) if possible, write the given general equation of a circle in standard form by completing the square, and identify the center and radius. graph the circle.

1. ( x^2 + 6x + y^2 - 2y = -1 )
2. ( x^2 + y^2 + 12y + 32 = 0 )
3. ( x^2 + 6x + y^2 - 2y + 3 = 0 )
4. ( x^2 - 4x + y^2 + 4y = -3 )
5. ( x^2 + 6x + y^2 + 8y + 9 = 0 )

Explanation:

Step1: Group \(x\) and \(y\) terms

Group the \(x\) - terms and \(y\) - terms together: \((x^{2}+6x)+(y^{2}-2y)= - 1\)

Step2: Complete the square for \(x\) - terms

For the \(x\) - terms \(x^{2}+6x\), use the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\). Here \(a = x\) and \(2ab=6x\), so \(b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\)

Step3: Complete the square for \(y\) - terms

For the \(y\) - terms \(y^{2}-2y\), use the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\). Here \(a = y\) and \(2ab = 2y\), so \(b = 1\). Then \(y^{2}-2y=(y - 1)^{2}-1\)

Step4: Substitute back into the equation

Substitute \((x + 3)^{2}-9+(y - 1)^{2}-1=-1\)

Step5: Simplify the equation

\((x + 3)^{2}+(y - 1)^{2}-10=-1\), then \((x + 3)^{2}+(y - 1)^{2}=9\)

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.

For the equation \((x + 3)^{2}+(y - 1)^{2}=9=(x-(-3))^{2}+(y - 1)^{2}=3^{2}\)

Answer:

The standard form of the circle is \((x + 3)^{2}+(y - 1)^{2}=9\), the center is \((-3,1)\) and the radius \(r = 3\)