Question

a circle has the equation 2(x - 2)^2+2y^2 = 2. (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 circle equation in standard form

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

Step2: Identify the center and radius

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

Answer:

(2,0)

For part (b), to graph the circle:

1. Plot the center at the point $(2,0)$.
2. Since the radius $r = 1$, from the center, move 1 unit up, down, left, and right to get points on the circle: $(3,0)$, $(1,0)$, $(2,1)$, $(2, - 1)$. Then draw a smooth curve passing through these points.

For part (c):

Step1: Find the x - intercepts

Set $y = 0$ in the equation $(x - 2)^2+y^2 = 1$. We get $(x - 2)^2=1$. Taking the square root of both sides, $x-2=\pm1$.

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

Step2: Find the y - intercepts

Set $x = 0$ in the equation $(x - 2)^2+y^2 = 1$. We have $(0 - 2)^2+y^2 = 1$, which simplifies to $4+y^2 = 1$, or $y^2=-3$. Since the square of a real number cannot be negative, there are no y - intercepts.

So the x - intercepts are $(1,0)$ and $(3,0)$, and there are no y - intercepts.