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 (1/2, - 3). (type an ordered pair, using integers or fractions.) the radius of the circle is 1/2. (type an integer or a fraction.) (b) use the graphing tool to graph the circle. (c) what are the intercepts? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the intercept(s) is/are (type an ordered pair. use a comma to separate answers as needed. type exact answers for each coordinate, using radicals as needed. simplify your answer.) b. there are no intercepts.

Explanation:

Step1: Find x - intercepts

Set \(y = 0\) in the circle equation \(x^{2}+y^{2}-x + 6y+9 = 0\). We get \(x^{2}-x + 9=0\). The discriminant of this quadratic equation \(ax^{2}+bx + c=0\) (where \(a = 1\), \(b=-1\), \(c = 9\)) is \(\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times9=1 - 36=- 35<0\). So, there are no real - valued x - intercepts.

Step2: Find y - intercepts

Set \(x = 0\) in the circle equation \(x^{2}+y^{2}-x + 6y+9 = 0\). We get \(y^{2}+6y + 9=0\). Factoring, we have \((y + 3)^{2}=0\). Solving for \(y\), we get \(y=-3\). The y - intercept is the point \((0,-3)\).

Answer:

A. \((0,-3)\)