Question

a circle has the equation x² + y² = 121. (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 (0,0). (type an ordered pair, using integers or decimals.) the radius of the circle is 11. (type an integer or a decimal.) (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 within 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.) b. there are no intercepts.

Explanation:

Step1: Recall circle - equation form

The standard form of a circle's equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}=121$, we can rewrite it as $(x - 0)^2+(y - 0)^2 = 11^2$. So the center $(h,k)=(0,0)$ and radius $r = 11$.

Step2: Find x - intercepts

Set $y = 0$ in the equation $x^{2}+y^{2}=121$. We get $x^{2}+0^{2}=121$, so $x^{2}=121$. Solving for $x$, we have $x=\pm11$. The x - intercepts are $(- 11,0)$ and $(11,0)$.

Step3: Find y - intercepts

Set $x = 0$ in the equation $x^{2}+y^{2}=121$. We get $0^{2}+y^{2}=121$, so $y^{2}=121$. Solving for $y$, we have $y=\pm11$. The y - intercepts are $(0,-11)$ and $(0,11)$.

Answer:

A. The intercept(s) is/are $(-11,0),(11,0),(0, - 11),(0,11)$