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 . (type an ordered pair, using integers or decimals.)

Explanation:

Step1: Rewrite the given equation in standard form

The standard form of a circle equation is $(x - h)^2+(y - k)^2=r^2$. Given $3(x - 3)^2+3y^2 = 3$, divide the entire equation by 3. We get $(x - 3)^2+y^2 = 1$.

Step2: Identify the center and radius

Comparing $(x - 3)^2+(y - 0)^2=1^2$ with the standard - form $(x - h)^2+(y - k)^2=r^2$, we can see that $h = 3$, $k = 0$, and $r = 1$.

Answer:

(a) $(3,0)$
(b) To graph the circle: Plot the center $(3,0)$. Then, from the center, move 1 unit up, down, left, and right to get points on the circle and draw a smooth curve passing through those points.
(c)

• x - intercepts: Set $y = 0$ in the equation $(x - 3)^2+y^2 = 1$. We have $(x - 3)^2=1$. Taking the square root of both sides, $x - 3=\pm1$. So $x=3 + 1=4$ or $x=3-1 = 2$. The x - intercepts are $(2,0)$ and $(4,0)$.
• 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.