Chapter 1, Problem 26RE

To determine

To find: the center and radius of each circle and graph the circle.

Answer to Problem 26RE

The center $\left(h,k\right)$ of the circle is $\left(1,0\right)$ , and the radius is 1.

  

Explanation of Solution

Given:

  $2x^2+2y^2-4x=0$

Calculation:

The standard form of an equation of a circle with radius r and canter $\left(h,k\right)$ is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Simplify the given expression. For this, divide each term by 2.

  $\begin{array}{l}\frac{2x^2}{2}+\frac{2y^2}{2}-\frac{4x}{2}=\frac{0}{2}\\ x^2+y^2-2x=0\end{array}$

In order to convert the given equation to standard form, complete the square in both x and y . Group the expressions involving x - and y -terms.

  $\left(x^2-2x\right)+y^2=0$

In order to complete the squares, take half the coefficient of x . Square the number and add it on both the sides of the equation.

  $\begin{array}{l}\left(x^2-2x+1\right)+y^2=0+1\\ \stackrel{\uparrow }{\overbrace{{\left(\frac{2}{2}\right)}^2=1}}\end{array}$

Factor the equation.

  ${\left(x-1\right)}^2+y^2=1^2$

The equation obtained can be rewritten as ${\left(x-1\right)}^2+{\left(y-0\right)}^2=1^2$ .

On comparing the equation with the standard form, the values are obtained areh as 1, k as 0, and r as 1.

Thus, the center $\left(h,k\right)$ of the circle is $\left(1,0\right)$ , and the radius is 1.

In order to graph the circle, first plot the center $\left(1,0\right)$ . Next, plot four points to complete the circle. Since the radius is 1, move 1 unit to the left from the center and plot the corresponding point. Similarly, plot points 1 unit to the right, up, and down from the center. Join the points using a smooth curve.

  

In order to find the x -intercepts, substitute 0 for y in the equation.

  $\begin{array}{l}{\left(x-1\right)}^2+0^2=1\\ {\left(x-1\right)}^2=1\end{array}$

Take square root on both the sides.

  $\begin{array}{l}\sqrt{{\left(x-1\right)}^2}=\sqrt{1}\\ x-1=\pm 1\end{array}$

Thus, $x-1=1$ or $x-1=-1$ .

Solve for x ,

  $\begin{array}{l}x-1=1\text{ or }x-1=-1\\ x=1+1\text{ or }x=-1+1\\ x=2\text{ or }x=0\end{array}$

Thus, the x -intercepts are 2 and 0.

In order to find the y -intercepts, substitute 0 for x in the equation.

  (0 - 1)2 + y2 = 1

  1 + y2 = 1

  $\begin{array}{l}{\left(0-1\right)}^2+y^2=1\\ 1+y^2=1\end{array}$

Solve for y .

Subtract 1 from each side.

  $\begin{array}{l}1+y^2-1=1-1\\ y^2=0\\ y=0\end{array}$

The y -intercept is 0.

Conclusion:

Therefore, the center $\left(h,k\right)$ of the circle is $\left(1,0\right)$ , and the radius is 1.The x -intercepts are 2 and 0. The y -intercept is 0.