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 . (type an integer or a decimal.)

Explanation:

Step1: Rewrite the given equation in standard form

The standard - form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Given the equation $3(x - 3)^2+3y^2 = 3$. Divide the entire equation by 3 to get it in standard form: $\frac{3(x - 3)^2}{3}+\frac{3y^2}{3}=\frac{3}{3}$, which simplifies to $(x - 3)^2+y^2 = 1$.

Step2: Identify the center and radius

Comparing $(x - 3)^2+(y - 0)^2 = 1$ with the standard - form $(x - h)^2+(y - k)^2=r^2$, we have $h = 3$, $k = 0$, and $r^2=1$. Taking the square root of both sides of $r^2 = 1$ (and considering $r>0$ since $r$ represents the radius of a circle), we get $r = 1$.

Answer:

(a) The center of the circle is $(3,0)$
The radius of the circle is $1$
(b) To graph the circle: Plot the center $(3,0)$. Then, from the center, move 1 unit up, 1 unit down, 1 unit left, and 1 unit right and plot these points. Connect these four points with a smooth curve to form the circle.
(c)

x - intercepts:

Set $y = 0$ in the equation $(x - 3)^2+y^2 = 1$. We get $(x - 3)^2+0^2 = 1$, so $(x - 3)^2=1$. Taking the square root of both sides, $x-3=\pm1$.
If $x - 3 = 1$, then $x=4$; if $x - 3=-1$, then $x = 2$. The x - intercepts are $(2,0)$ and $(4,0)$.

y - intercepts:

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