Chapter 1.4, Problem 64AYU

(a)

To determine

To find: the radius of the circle

Answer to Problem 64AYU

The radius $r$ is the fixed distance from a fixed point (center) to all the points lay on the circle.

Explanation of Solution

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

The radius $r$ is the fixed distance from a fixed point (center) to all the points lay on the circle.

Conclusion:

The radius $r$ is the fixed distance from a fixed point (center) to all the points lay on the circle.

(b)

To determine

To find: the equation of the circle

Answer to Problem 64AYU

Thus theCartesian plane is;

  

The equation of a circle is the equation of the circle with center $\left(1,3\right)$ is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=r^2$

Explanation of Solution

Given:Center is at $\left(1,3\right)$

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

The graph of a circle whose center is at $\left(1,3\right)$

  

The equation of the circle with center $\left(1,3\right)$ is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=r^2$

Conclusion:

Thus theCartesian plane is;

  

The equation of a circle is the equation of the circle with center $\left(1,3\right)$ is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=r^2$

(c)

To determine

To find: the equation of the circle

Answer to Problem 64AYU

Thus theCartesian plane is;

  

The equation of the circle with center $\left(-1,3\right)$ is ${\left(x+1\right)}^2+{\left(y-3\right)}^2=r^2$

Explanation of Solution

Given:Center is at $\left(-1,3\right)$ .

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

The graph of a circle whose center is at $\left(-1,3\right)$ .

  

The equation of the circle with center $\left(-1,3\right)$ is ${\left(x+1\right)}^2+{\left(y-3\right)}^2=r^2$

Conclusion:

Thus theCartesian plane is;

  

The equation of the circle with center $\left(-1,3\right)$ is ${\left(x+1\right)}^2+{\left(y-3\right)}^2=r^2$

(d)

To determine

To find: the equation of the circle

Answer to Problem 64AYU

Thus theCartesian plane is;

  

The equation of the circle with center $\left(-1,-3\right)$ is ${\left(x+1\right)}^2+{\left(y+3\right)}^2=r^2$

Explanation of Solution

Given:Center is at $\left(-1,-3\right)$

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

The graph of a circle whose center is at $\left(-1,-3\right)$ :

  

The equation of the circle with center $\left(-1,-3\right)$ is ${\left(x+1\right)}^2+{\left(y+3\right)}^2=r^2$

Conclusion:

Thus theCartesian plane is;

  

The equation of the circle with center $\left(-1,-3\right)$ is ${\left(x+1\right)}^2+{\left(y+3\right)}^2=r^2$

(e)

To determine

To find: the equation of the circle

Answer to Problem 64AYU

Thus theCartesian plane is;

  

The equation of the circle with center $\left(1,-3\right)$ is ${\left(x-1\right)}^2+{\left(y+3\right)}^2=r^2$

Explanation of Solution

Given:Center is at $\left(1,-3\right)$

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

The graph of a circle whose center is at $\left(1,-3\right)$

  

The equation of the circle with center $\left(1,-3\right)$ is ${\left(x-1\right)}^2+{\left(y+3\right)}^2=r^2$

Conclusion:

Thus theCartesian plane is;

  

The equation of the circle with center $\left(1,-3\right)$ is ${\left(x-1\right)}^2+{\left(y+3\right)}^2=r^2$

(f)

To determine

To explain: the role of the center of the circle in the equation of the circle.

Answer to Problem 64AYU

These four points

  $\left(h-r,k\right),\left(h+r,k\right),\left(h,k+r\right),\left(h,k-r\right)$ can be used as guides to obtain the graph.

Explanation of Solution

Explanations:

The standard equation of a circle with center $\left(h,k\right)$ and radius $4$ is:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{............}(1)$

Thecenter of a circle leads an important role in the equation of the circle. For a circle, find those points $\left(x,y\right)$ which are at a fixed distance (radius) say $r$ from the center say $\left(h,k\right)$ . If $\left(h,k\right)$ is changed, the equation of the circle will get changed.

For a particular circle with center $\left(h,k\right)$ and radius $r$ , there will be infinite number of points $\left(x,y\right)$ satisfying the circle equation ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

To draw the graph of a circle $\left(h,k\right)$ shows an important role. First plot the center $\left(h,k\right)$ . Since the radius is $r$ , locate four points on the circle by plotting points $r$ units to the right, up and down from the center.

These four points $\left(h-r,k\right),\left(h+r,k\right),\left(h,k+r\right),\left(h,k-r\right)$ can be used as guides to obtain the graph.

Conclusion:

Thus, these four points $\left(h-r,k\right),\left(h+r,k\right),\left(h,k+r\right),\left(h,k-r\right)$ can be used as guides to obtain the graph.