Question about ixl.com

Write the equation in standard form for the circle passing through ( - 6,2) centered at the origin

Ad

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

Posted on Apr 04, 2015

Ad

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

Ad

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

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

Good luck.

Paul

Apr 15, 2015 | Office Equipment & Supplies

Assuming the 'standard form' is "slope-intercept", calculate the slope from the equation m = __y2-y1__ =__ 5 - 1__ = __ 4__ = -2

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

**y = (-2) x + 13**

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

Oct 10, 2014 | Computers & Internet

Calcualte the slope of the line as

a=(7-(-11))/(10-(-5))=18/15=6/5

Use the fact that the line passes through one of the two points, for example (10,7)

7=(6/5)*10+b=12+b

Obtain b as b=7-12=-5

The equation of the line in functional form is y=(6/5)x-5

Multiply everything by 5 to clear the fraction

5y=6x-25 or 0=6x-5y-25

Finally, the equation in general form (standard?) is**6x-5y-25=0**.

Check the calculation by verifying that the point (10,7) lies on the line.

6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!

Check that the second point (-5,-11) lies on the line also (if you want to)

6*(-5)-5*(-11)-25=-30+55-25=0

That checks OK.

a=(7-(-11))/(10-(-5))=18/15=6/5

Use the fact that the line passes through one of the two points, for example (10,7)

7=(6/5)*10+b=12+b

Obtain b as b=7-12=-5

The equation of the line in functional form is y=(6/5)x-5

Multiply everything by 5 to clear the fraction

5y=6x-25 or 0=6x-5y-25

Finally, the equation in general form (standard?) is

Check the calculation by verifying that the point (10,7) lies on the line.

6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!

Check that the second point (-5,-11) lies on the line also (if you want to)

6*(-5)-5*(-11)-25=-30+55-25=0

That checks OK.

Dec 04, 2011 | Super Tutor Pre Algebra (ESDPALG)

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...

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)

2. Quadratic Equations y= ax^2+bx+c

3. Exponential Equations y= ab^x

4. Cubic Equations y=ax^3+ bx^2+cx+d

5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e

6. Equation of a circle (x-h)^2+(y-k)^2= r^2

7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).

8. Proportional equations y=kx; y= k/x, etc.

Jun 14, 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

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

Apr 14, 2014 | ixl.com

101 people viewed this question

Usually answered in minutes!

×