Question

a circle has the equation $x^{2}+y^{2}-x + 6y+9 = 0$. (a) find the center (h,k) and radius r of the circle. (b) graph the circle. (c) find the intercepts, if any, of the graph. (a) the center of the circle is $(\frac{1}{2},-3)$. (type an ordered pair, using integers or fractions.) the radius of the circle is $\square$. (type an integer or a fraction.)

Explanation:

Step1: Rewrite the circle equation in standard form

The general equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Given $x^{2}+y^{2}-x + 6y+9 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}-x=(x-\frac{1}{2})^{2}-\frac{1}{4}$. For the $y$ - terms: $y^{2}+6y=(y + 3)^{2}-9$. So the equation becomes $(x-\frac{1}{2})^{2}-\frac{1}{4}+(y + 3)^{2}-9 + 9=0$, which simplifies to $(x-\frac{1}{2})^{2}+(y + 3)^{2}=\frac{1}{4}$.

Step2: Identify the radius

Comparing $(x-\frac{1}{2})^{2}+(y + 3)^{2}=\frac{1}{4}$ with the standard - form $(x - h)^2+(y - k)^2=r^2$, we have $r^{2}=\frac{1}{4}$. Taking the square root of both sides, $r=\frac{1}{2}$ (we take the positive value since the radius is non - negative).

Answer:

$\frac{1}{2}$