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.
$x^{2}-6x + y^{2}-16 = 0$
the equation in standard form is
(simplify your answer.)

Explanation:

Step1: Complete the square for x - terms

The coefficient of $x$ is $- 6$. Half of it is $\frac{-6}{2}=-3$, and its square is $(-3)^2 = 9$. Add 9 to both sides of the equation $x^{2}-6x + y^{2}-16=0$.
$x^{2}-6x + 9+y^{2}-16=9$.

Step2: Rewrite the left - hand side as perfect squares

Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, we can rewrite $x^{2}-6x + 9$ as $(x - 3)^2$. So the equation becomes $(x - 3)^2+y^{2}=25$.

Answer:

$(x - 3)^2+y^{2}=25$