Question

in problems 23 - 36, (a) find the center (h,k) and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any. 23. $x^{2}+y^{2}=4$ 24. $x^{2}+(y - 1)^{2}=1$ 25. $2(x - 3)^{2}+2y^{2}=8$

Explanation:

For the circle equation \(2(x - 3)^{2}+2y^{2}=8\)

(a) Find the center \((h,k)\) and radius \(r\)

Step1: Rewrite the equation in standard form

First, divide the entire equation \(2(x - 3)^{2}+2y^{2}=8\) by 2. We get \((x - 3)^{2}+y^{2}=4\). 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.
$$(x - 3)^{2}+(y - 0)^{2}=2^{2}$$

Step2: Identify the center and radius

Comparing with the standard form, we can see that \(h = 3\), \(k = 0\) and \(r=2\).

(b) Graph the circle

To graph the circle, plot the center \((3,0)\) on the coordinate plane. Then, using the radius \(r = 2\), we can find points on the circle by moving 2 units up, down, left and right from the center.

(c) Find the intercepts

x - intercepts:

Set \(y = 0\) in the equation \((x - 3)^{2}+y^{2}=4\).

Step1: Substitute \(y = 0\)

When \(y = 0\), the equation becomes \((x - 3)^{2}=4\).

Step2: Solve for \(x\)

Take the square - root of both sides: \(x-3=\pm2\).
If \(x - 3=2\), then \(x=5\); if \(x - 3=-2\), then \(x = 1\). So the x - intercepts are \((1,0)\) and \((5,0)\).

y - intercepts:

Set \(x = 0\) in the equation \((x - 3)^{2}+y^{2}=4\).

Step1: Substitute \(x = 0\)

When \(x = 0\), we have \((0 - 3)^{2}+y^{2}=4\), which simplifies to \(9 + y^{2}=4\), or \(y^{2}=-5\). Since the square of a real number cannot be negative, there are no y - intercepts.

Answer:

(a) Center \((h,k)=(3,0)\), radius \(r = 2\)
(b) Graph with center at \((3,0)\) and radius 2
(c) x - intercepts: \((1,0)\) and \((5,0)\); y - intercepts: None