Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation. x² - 6x + y² - 7 = 0 the equation in standard form is □. (simplify your answer.)

Explanation:

Step1: Group the x - terms and move the constant

Group \(x^{2}-6x\) together and move the constant term to the other side of the equation.
\(x^{2}-6x+y^{2}=7\)

Step2: Complete the square for the x - terms

For the quadratic expression \(ax^{2}+bx + c\) (here \(a = 1\), \(b=-6\) for the \(x\) - part), the formula for completing the square is \((x+\frac{b}{2})^{2}-\frac{b^{2}}{4}\).
For \(x^{2}-6x\), we have \(\frac{b}{2}=\frac{-6}{2}=-3\), and \((x - 3)^{2}=x^{2}-6x + 9\). So we add 9 to both sides of the equation:
\(x^{2}-6x+9+y^{2}=7 + 9\)

Step3: Write 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.
\((x - 3)^{2}+y^{2}=16\)

Answer:

The equation in standard form is \((x - 3)^{2}+y^{2}=16\). The center of the circle is \((3,0)\) and the radius \(r = 4\).