Question

a circle has the equation 3(x - 3)^2 + 3y^2 = 3. (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 (3,0). (type an ordered pair, using integers or decimals.) the radius of the circle is 1. (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: Rewrite the circle equation

Given $3(x - 3)^2+3y^2 = 3$, divide through by 3 to get $(x - 3)^2+y^2 = 1$. The standard - form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Here $h = 3,k = 0,r = 1$.

Step2: Find x - intercepts

Set $y = 0$ in the equation $(x - 3)^2+y^2 = 1$. Then $(x - 3)^2=1$. Taking the square - root of both sides, we have $x−3=\pm1$.

• For $x - 3 = 1$, $x=4$.
• For $x - 3=-1$, $x = 2$. So the x - intercepts are $(2,0)$ and $(4,0)$.

Step3: Find y - intercepts

Set $x = 0$ in the equation $(x - 3)^2+y^2 = 1$. Then $(0 - 3)^2+y^2 = 1$, which simplifies to $9 + y^2=1$, or $y^2=-8$. Since the square of a real number cannot be negative, there are no y - intercepts.

Answer:

(a) Center: $(3,0)$; Radius: $1$
(b) Cannot be shown in this text - based format. Use a graphing utility (e.g., Desmos, Geogebra) to graph the circle with center $(3,0)$ and radius $1$.
(c) A. $(2,0),(4,0)$