\( \leftarrow \) A circle has the equation \( x^{2}+y^{2}-3 x-6 y+9=0 \). (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 find the center and radius of the circle given by the equation \( x^{2}+y^{2}-3 x-6 y+9=0 \), we first rearrange the equation to standard form by completing the square. Starting with the equation: \[ x^{2}-3x + y^{2}-6y + 9 = 0 \] We can rewrite it as: \[ x^{2}-3x + y^{2}-6y = -9 \] Next, we complete the square for \( x \) and \( y \): 1. For \( x^{2}-3x \): \[ = (x - \frac{3}{2})^{2} - \frac{9}{4} \] 2. For \( y^{2}-6y \): \[ = (y - 3)^{2} - 9 \] Substituting these back into the equation gives: \[ \left( x - \frac{3}{2} \right)^{2} - \frac{9}{4} + \left( y - 3 \right)^{2} - 9 = -9 \] This simplifies to: \[ \left( x - \frac{3}{2} \right)^{2} + \left( y - 3 \right)^{2} = \frac{9}{4} \] Now we identify the center \((h, k)\) and the radius \(r\): - The center is \(\left( \frac{3}{2}, 3 \right)\) - The radius \(r\) is \(\sqrt{\frac{9}{4}} = \frac{3}{2}\) So, the center of the circle is: \[ \left( \frac{3}{2}, 3 \right) \] Now, for the intercepts: - The x-intercepts occur when \( y = 0 \): \[ \left( x - \frac{3}{2} \right)^{2} + \left( 0 - 3 \right)^{2} = \frac{9}{4} \] Solving this gives two solutions. - The y-intercepts occur when \( x = 0 \): \[ \left( 0 - \frac{3}{2} \right)^{2} + \left( y - 3 \right)^{2} = \frac{9}{4} \] Similar processing will yield the y-intercepts. Both intercepts can be calculated to find their precise values!