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

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

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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.

Posted on Apr 04, 2015


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Write the equation in standard form for the circle with radius 1 centered at the origin.

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.


Apr 15, 2015 | Office Equipment & Supplies

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Write an equation in standard form for vertex(6,1) passing through the point (4,5)

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

Oct 10, 2014 | Computers & Internet

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Write an equation of the line in standard form that pass through (-5,-11) and 10,7)

Calcualte the slope of the line as
Use the fact that the line passes through one of the two points, for example (10,7)
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)
That checks OK.

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

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Find the radius of the equation x^2+y^2-2x-2y=14

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).

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To graph a circle with radius 4

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)

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

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What are the equations and its kinds?


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

1 Answer


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.

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

1 Answer

How can I get my calculator to highlight more than one function, to graph a circle for example?


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

1 Answer

Analytic geometry

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


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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