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Now if I give you the answer how are you going to learn anything about this? Your only asking for a flowchart. I will give you a hint. break the problem down. All the information needed except the Formulas used is in your post.

Posted on Sep 15, 2010

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Posted on Jan 02, 2017

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(pi)R squared (Fixya won't let me use the sign for Pi)

Area of a circle=pi*radius^2 or pi times radius times radius

example: suppose the radius of a circle is 10 inches, then the Area of the circle would be pi times 10 times 10 or 100*pi

pi represents the circumference (distance around) of a circle / diameter and for any circle is constantly the same number which is approximately 3.14

so in my example 100*pi=100*3.14 or 314 square inches

Area of a circle=pi*radius^2 or pi times radius times radius

example: suppose the radius of a circle is 10 inches, then the Area of the circle would be pi times 10 times 10 or 100*pi

pi represents the circumference (distance around) of a circle / diameter and for any circle is constantly the same number which is approximately 3.14

so in my example 100*pi=100*3.14 or 314 square inches

Jul 03, 2014 | Computers & Internet

A circle is the set of points equidistant from a given point, so

(x-a)^2 + (y-b)^2 = c will define a circle, where a and b are the coordinates of the center and c is the square of the radius.

(x-a)^2 + (y-b)^2 = c will define a circle, where a and b are the coordinates of the center and c is the square of the radius.

Aug 23, 2013 | Casio Fx300-ms Scientific Calculator With...

The radius and the coordinates of the center will completely define a circle. Some other measurements include the diameter, circumference, and area. Then there are parts of the circle, such as arc and chord.

Aug 23, 2013 | Casio Fx300-ms Scientific Calculator With...

Press [2nd][PGRM] to open the (DRAW) utility. Scroll down to reach the line 9: Circle( and press ENTER. Complete the command by supplying the coordinates of the circle center and the radius.

Circle(0,0,4) draws a circle with center at the origin (0,0) and radius 4.

By the way the equation of a circle is not a function. You have to cut it into the upper brach and the lower part to graph it.

y=+SQRT(16-x^2) and y=-SQRT(16-x^2)

Circle(0,0,4) draws a circle with center at the origin (0,0) and radius 4.

By the way the equation of a circle is not a function. You have to cut it into the upper brach and the lower part to graph it.

y=+SQRT(16-x^2) and y=-SQRT(16-x^2)

Sep 01, 2011 | Texas Instruments TI-84 Plus Silver...

Create a flowchart that computes for the circumference of circle. Assume that the value for radius is 7.23.Your flowchart should be able to show the value of the circumference.The pie value is 3.1416 which is constant and never be changed.You can compute the circumference by using this formula 2piR?

Jul 06, 2011 | Computers & Internet

Use the DRAW program.

Press [2nd][PRGM] (DRAW)

Select [9:Circle (]

Command echoes on main screen

Complete the command with coordinates of the center and the radiius

Close parenthesis and press [ENTER]

The circle is displayed.

Ex: Circle(0,3,2) draw a circle with cntere at (0,3) and radius equal to 2.

Alternatively: If the conics application is on the calculator run it.

Press [APPS]

Select [Conics]

Select [1:CIRCLE]

Choose the canonical form (X-H)^2 + (Y-K)^2=R^2 where H, and K are the coordinates of the center and R is the radius.

You will be prompted to enter the values for H,K, and R

You can also use the general equation for a circle aX^2 +aY^2+BX+CY +D=0

Enter the values of a, B, C and D.Change the values to get what you want.

Press [2nd][PRGM] (DRAW)

Select [9:Circle (]

Command echoes on main screen

Complete the command with coordinates of the center and the radiius

Close parenthesis and press [ENTER]

The circle is displayed.

Ex: Circle(0,3,2) draw a circle with cntere at (0,3) and radius equal to 2.

Alternatively: If the conics application is on the calculator run it.

Press [APPS]

Select [Conics]

Select [1:CIRCLE]

Choose the canonical form (X-H)^2 + (Y-K)^2=R^2 where H, and K are the coordinates of the center and R is the radius.

You will be prompted to enter the values for H,K, and R

You can also use the general equation for a circle aX^2 +aY^2+BX+CY +D=0

Enter the values of a, B, C and D.Change the values to get what you want.

Apr 01, 2010 | Texas Instruments TI-83 Plus Calculator

30" diameter? Radius is 15" (half of diameter).

30" circumference (distance around) is a little more complicated...

The number is the ratio of the circumference of a circle to the diameter. The value of is approximately 3.14159265358979323846...The diameter of a circle is twice the radius. Given the diameter or radius of a circle, we can find the circumference. We can also find the diameter (and radius) of a circle given the circumference. The formulas for diameter and circumference of a circle are listed below. We round to 3.14 in order to simplify our calculations.

30" circumference (distance around) is a little more complicated...

The number is the ratio of the circumference of a circle to the diameter. The value of is approximately 3.14159265358979323846...The diameter of a circle is twice the radius. Given the diameter or radius of a circle, we can find the circumference. We can also find the diameter (and radius) of a circle given the circumference. The formulas for diameter and circumference of a circle are listed below. We round to 3.14 in order to simplify our calculations.

Nov 17, 2009 | Gateway FPD1975W 19" LCD Monitor

The Circle Class Using Data Hiding and Encapsulation
package shapes; // Specify a package for the class
public class Circle { // The class is still public
// This is a generally useful constant, so we keep it public
public static final double PI = 3.14159;
protected double r; // Radius is hidden, but visible to subclasses
// A method to enforce the restriction on the radius
// This is an implementation detail that may be of interest to subclasses
protected checkRadius(double radius) {
if (radius < 0.0)
throw new IllegalArgumentException("radius may not be negative.");
}
// The constructor method
public Circle(double r) {
checkRadius(r);
this.r = r;
}
// Public data accessor methods
public double getRadius() { return r; };
public void setRadius(double r) {
checkRadius(r);
this.r = r;
}
// Methods to operate on the instance field
public double area() { return PI * r * r; }
public double circumference() { return 2 * PI * r; }
}

Mar 25, 2009 | Computers & Internet

assuming the question is what is the circle equation?

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

Nice try!

If nobody has helped you since September 2008, when you asked the question, you probably scored a "zero" in your computer-programming course, and probably don't need an answer to your "homework" assignment.

If nobody has helped you since September 2008, when you asked the question, you probably scored a "zero" in your computer-programming course, and probably don't need an answer to your "homework" assignment.

Sep 15, 2008 | Computers & Internet

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