Question

a circle has the equation 4(x - 3)^2+4y^2 = 4. (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.

Explanation:

Step1: Rewrite the equation in standard form

Divide the given equation $4(x - 3)^2+4y^2 = 4$ by 4 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.

Step2: Identify the center and radius

Comparing $(x - 3)^2+y^2 = 1$ with $(x - h)^2+(y - k)^2=r^2$, we have $h = 3$, $k = 0$, and $r = 1$.

Step3: Find the 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, $x - 3=\pm1$. So $x=3 + 1=4$ or $x=3 - 1=2$.

Step4: Find the 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) To graph the circle, plot the center at the point $(3,0)$ and then draw a circle with a radius of 1 unit around it.
(c) x - intercepts: $x = 2$ and $x = 4$, y - intercepts: None