Question

1. if the equation of a circle is (x^{2}+y^{2}-6x + 4y+9 = 0), what is its center?

Explanation:

Step1: Rewrite the equation in standard form.

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle. Given $x^{2}+y^{2}-6x + 4y+9 = 0$. Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$.
For the $y$ - terms: $y^{2}+4y=(y + 2)^{2}-4$.
So the equation becomes $(x - 3)^{2}-9+(y + 2)^{2}-4 + 9=0$.

Step2: Simplify the equation.

$(x - 3)^{2}+(y + 2)^{2}-4=0$, then $(x - 3)^{2}+(y + 2)^{2}=4$.

Answer:

$(3,-2)$