Question about Office Equipment & Supplies

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The formula for a circle in standard form in (x-h)^2 + (y-k)^2 = r^2, where r is the radius, the point (h,k) is the centre of the circle, and ^2 means squared.

Substituting in 0 for h and 0 for k, and 1 for r, you will get your formula.

Good luck.

Paul

Posted on Apr 15, 2015

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Posted on Jan 02, 2017

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A. The equation of a circle is (x-a)^2 + (y-b)^2 = r^2, with (a,b) being the centre of the circle and r being the radius.

Good luck,

Paul

Good luck,

Paul

May 01, 2015 | Office Equipment & Supplies

The equation of a circle is (x-h)^2 + (y-k)^2 = r^, where h and k are the x and y coordinates of the centre of the circle. However the centre is the origin, so we have x^2 + y^2 = r^2.

Now, we need to figure out r. However, we can calculate r because it is the distance from the origin to (-6,2). We can use Pythagorean Theorum, a^2 + b^2 = c^2, were a is -6 and b is 2. We get 6^2 + 2^2 =c^2. c^2= 40.

Thus, the formula of the circle is x^2 + y^2 = 40.

Good luck.

Paul

Now, we need to figure out r. However, we can calculate r because it is the distance from the origin to (-6,2). We can use Pythagorean Theorum, a^2 + b^2 = c^2, were a is -6 and b is 2. We get 6^2 + 2^2 =c^2. c^2= 40.

Thus, the formula of the circle is x^2 + y^2 = 40.

Good luck.

Paul

Apr 14, 2014 | ixl.com

Try casting the equation of the circle into the form (x-h)^2+(y-k)^2=R^2. Here (h,k) would be the center and R the radius.

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

Sep 20, 2011 | SoftMath Algebrator - Algebra Homework...

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...

The equation of a circle with radius 2 and centered on (-1, -2) is

(x+1)^2 + (y+2)^2 -4 = 0

(x+1)^2 + (y+2)^2 -4 = 0

Oct 17, 2010 | Cameras

how big is the triangle? if the coins are a mile in diameter doesnt mean you cant have an equilateral triangle with legs of 32 miles long.

or are you asking for the equation?

i will send you on the path to the equation.

if you draw a circle on a piece of paper then draw a triangle around it where is the center? if the triangle were a perfect equilateral then the circle would be a perfect circle touching the sides of the triangle.

its important to know where the center is because the center of the circle to the edge of the circle is the radius.

so the center of a triangle is the center of the circle. therfeore the radius of the circle will equal hieght/2...

thinking that way you can say that if there were three coins then the center of the triangle cant be occupied by a coin but is very close to an edge. where is the center of the coin?

or are you asking for the equation?

i will send you on the path to the equation.

if you draw a circle on a piece of paper then draw a triangle around it where is the center? if the triangle were a perfect equilateral then the circle would be a perfect circle touching the sides of the triangle.

its important to know where the center is because the center of the circle to the edge of the circle is the radius.

so the center of a triangle is the center of the circle. therfeore the radius of the circle will equal hieght/2...

thinking that way you can say that if there were three coins then the center of the triangle cant be occupied by a coin but is very close to an edge. where is the center of the coin?

Jun 10, 2010 | Mathsoft StudyWorks! Middle School Deluxe...

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

Hello,

Suppose you want to draw the circle the equation of which is X^2+Y^2=4, a circle centered at the origin with radius 2.

When you isolate Y^2 and then extarct the squre root you obtain two solutions, call them Y1 and Y2

Y1=SQRT(4-X^2), Y2= -SQRT(4-X^2)

You press[Y=] and type in the right hand side of the equations above, one in Y1= and the other in Y2. Press [GRAPH].

Ih a graphic does not appear within the scree window adjust the window parameters ( [WINDOW]) or Press [ZOOM] and select ZoomIn, ZoomOut, ZoomSTD, etc.

Explore the program Draw accessible by pressing [2nd][DRAW].

Hope it helps

Suppose you want to draw the circle the equation of which is X^2+Y^2=4, a circle centered at the origin with radius 2.

When you isolate Y^2 and then extarct the squre root you obtain two solutions, call them Y1 and Y2

Y1=SQRT(4-X^2), Y2= -SQRT(4-X^2)

You press[Y=] and type in the right hand side of the equations above, one in Y1= and the other in Y2. Press [GRAPH].

Ih a graphic does not appear within the scree window adjust the window parameters ( [WINDOW]) or Press [ZOOM] and select ZoomIn, ZoomOut, ZoomSTD, etc.

Explore the program Draw accessible by pressing [2nd][DRAW].

Hope it helps

Sep 04, 2009 | Texas Instruments TI-84 Plus Calculator

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

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