A circle has the equation \( x^{2}+y^{2}-x-6 y+9=0 \) (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 (Type an ordered pair, using integers or fractions.)

To solve the problem, we will follow the steps outlined: ### Part (a): Find the center \((h, k)\) and radius \(r\) of the circle. The equation of the circle is given as: \[ x^{2} + y^{2} - x - 6y + 9 = 0 \] We can rearrange this equation into the standard form of a circle, which is: \[ (x - h)^{2} + (y - k)^{2} = r^{2} \] To do this, we will complete the square for both \(x\) and \(y\). 1. **Rearranging the equation:** \[ x^{2} - x + y^{2} - 6y + 9 = 0 \] Move the constant to the other side: \[ x^{2} - x + y^{2} - 6y = -9 \] 2. **Completing the square for \(x\):** Take the coefficient of \(x\) (which is \(-1\)), halve it to get \(-\frac{1}{2}\), and square it to get \(\frac{1}{4}\). \[ x^{2} - x = \left(x - \frac{1}{2}\right)^{2} - \frac{1}{4} \] 3. **Completing the square for \(y\):** Take the coefficient of \(y\) (which is \(-6\)), halve it to get \(-3\), and square it to get \(9\). \[ y^{2} - 6y = (y - 3)^{2} - 9 \] 4. **Substituting back into the equation:** \[ \left(x - \frac{1}{2}\right)^{2} - \frac{1}{4} + (y - 3)^{2} - 9 = -9 \] Simplifying gives: \[ \left(x - \frac{1}{2}\right)^{2} + (y - 3)^{2} = -9 + 9 + \frac{1}{4} \] \[ \left(x - \frac{1}{2}\right)^{2} + (y - 3)^{2} = \frac{1}{4} \] 5. **Identifying the center and radius:** From the standard form \((x - h)^{2} + (y - k)^{2} = r^{2}\), we can see that: - The center \((h, k)\) is \(\left(\frac{1}{2}, 3\right)\) - The radius \(r\) is \(\sqrt{\frac{1}{4}} = \frac{1}{2}\) Thus, the center of the circle is: \[ \left(\frac{1}{2}, 3\right) \] ### Part (b): Graph the circle. To graph the circle, we will plot the center \(\left(\frac{1}{2}, 3\right)\) and draw a circle with radius \(\frac{1}{2}\). ### Part (c): Find the intercepts, if any, of the graph. To find the intercepts, we will set \(y = 0\) for the x-intercepts and \(x = 0\) for the y-intercepts. 1. **Finding x-intercepts:** Set \(y = 0\) in the original equation: \[ x^{2} - x + 0^{2} - 6(0) + 9 = 0 \] This simplifies to: \[ x^{2} - x + 9 = 0 \] We can use the quadratic formula to find the roots: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^{2} - 4(1)(9)}}{2(1)} = \frac{1 \pm \sqrt{1 - 36}}{2} = \frac{1 \pm \sqrt{-35}}{2} \] Since the discriminant is negative, there are no real x-intercepts. 2. **Finding y-intercepts:** Set \(x = 0\) in the original equation: \[ 0^{2} + y^{2} - 0 - 6y + 9 = 0 \] This simplifies to: \[ y^{2} - 6y + 9 = 0 \] Factoring gives: \[ (y - 3)^{2} = 0 \] Thus, \(y = 3\) is a double root, meaning there is one y-intercept at \((0, 3)\). ### Summary of Results: - The center of the circle is \(\left(\frac{1}{2}, 3\right)\). - The radius of the circle is \(\frac{1}{2}\). - There are no x-intercepts, and the y-intercept is at \((0, 3)\).