Question

given a circle with the equation (x^{2}+y^{2}-6x - 2y+5 = 0), what are the coordinates of its center and the length of its radius? use the keypad to enter your answers in the boxes. center ( , ) radius

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 and $r$ is the radius. Given $x^{2}+y^{2}-6x - 2y+5 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$. For the $y$ - terms: $y^{2}-2y=(y - 1)^{2}-1$. So the equation becomes $(x - 3)^{2}-9+(y - 1)^{2}-1 + 5=0$.

Step2: Simplify the equation

Rearrange the equation: $(x - 3)^{2}+(y - 1)^{2}=9 + 1-5$.

Step3: Calculate the radius

$(x - 3)^{2}+(y - 1)^{2}=5$. Comparing with the standard form $(x - a)^{2}+(y - b)^{2}=r^{2}$, we have $r=\sqrt{5}$, and the center $(a,b)=(3,1)$.

Answer:

center (3, 1)
radius $\sqrt{5}$