Review the standard and expanded forms of circle equations, and solve problems concerning them.

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Ravengal101

8 years agoPosted 8 years ago. Direct link to Ravengal101's post “What should be done to ad...”

What should be done to adjust the radius of the circle in the practice? I could move the centre, but I'm lost as to how the radius should be adjusted.

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(28 votes)

Timothy Soh

8 years agoPosted 8 years ago. Direct link to Timothy Soh's post “Move your cursor outwards...”

Move your cursor outwards slowly until the center of the circle is minimized. Then you can click and drag to adjust the radius of the circle.

It is a bit tricky when the circle is small but it can be done(62 votes)

M. McManus

7 years agoPosted 7 years ago. Direct link to M. McManus's post “Parabolas are called Quad...”

Parabolas are called Quadratics. What are circle equations called? Are they also called Quadratics because of the square? Thanks.

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(7 votes)

kubleeka

7 years agoPosted 7 years ago. Direct link to kubleeka's post “Parabolas, hyperbolas, el...”

Parabolas, hyperbolas, ellipses, and circles (which are just special ellipses) are collectively called "Conic Sections", since you can find each of these curves as the intersection between an infinite cone and a plane.

"Quadratic" refers to a polynomial of degree 2, and hence only describes parabolas.

Equations that describe circles don't really have a special name.

(35 votes)

1262076

7 years agoPosted 7 years ago. Direct link to 1262076's post “In problem 2.2 how did yo...”

In problem 2.2 how did you get √17?

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(7 votes)

Judith Gibson

7 years agoPosted 7 years ago. Direct link to Judith Gibson's post “I think the graph is dece...”

I think the graph is deceiving in that it is hard to read accurately.

At first glance it looks like the circle goes though the point (9,0) and therefore has a radius of 4.

However, upon closer looking, the circumference of the circle is just a bit beyond (9,0) and actually hits the point (9,-1).

That's why (using the Pythagorean theorem) the distance for the radius is calculated between the center (5,0) and the point (9,-1) which lies on its circumference. (Note that the point (9,1) could have been used instead as could (1,1) or (1,-1).)(26 votes)

arthurgrayuk

4 years agoPosted 4 years ago. Direct link to arthurgrayuk's post “How can you find square r...”

How can you find square roots without a calculator?

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(5 votes)

KLaudano

4 years agoPosted 4 years ago. Direct link to KLaudano's post “If you don't need an exac...”

If you don't need an exact answer, you can use linear approximations. Suppose we are trying to find the square root of a number, x, and we know the square root of a number, k, that is "close" to x. (k will likely be perfect square as it makes the math easier). The square root of x is approximately equal to (x + k)/(2 * sqrt(k)).

(9 votes)

Louise

7 years agoPosted 7 years ago. Direct link to Louise's post “Hi! I really need an urge...”

Hi! I really need an urgent response towards how should I find the equation of circle if the only given were the center (-1,-7) and a point (2,11).

Well, I tried doing using the standard equation of the circle and imputed the given sample and tried to square it. Unfortunately, WileyPLUS (Its a book converted into sort of online stuffs and other than pen and paper activities its all made online) as part of our school marked my answers wrong... can anyone help me? I really need to perfect it.

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(2 votes)

Judith Gibson

7 years agoPosted 7 years ago. Direct link to Judith Gibson's post “The standard equation for...”

The standard equation for a circle centred at (h,k) with radius r

is (x-h)^2 + (y-k)^2 = r^2

So your equation starts as ( x + 1 )^2 + ( y + 7 )^2 = r^2

Next, substitute the values of the given point (2 for x and 11 for y), getting

3^2 + 18^2 = r^2,

so r^2 = 333.

The final equation is (x+1)^2 + (y+7)^2 = 333

Hope this helps!(12 votes)

wairishtinah

7 years agoPosted 7 years ago. Direct link to wairishtinah's post “Find the centre and radiu...”

Find the centre and radius of a circle whose equation is x2+y2-4y-21

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(4 votes)

Judith Gibson

7 years agoPosted 7 years ago. Direct link to Judith Gibson's post “What you have written isn...”

What you have written isn't an equation because it has no equals sign, so presumably you mean x^2 + y^2 - 4y - 21 = 0

We know that the general equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius.

So add 21 to both sides to get the constant term to the righthand side of the equation.

x^2 + y^2 -4y = 21

Then complete the square for the y terms.

x^2 + y^2 - 4y + 4 = 21 + 4

Then factor.

x^2 + ( y - 2 )^2 = 5^2

So, the center is at ( 0, 2 ) and the radius is 5.(5 votes)

sheeraz.mohammd

3 years agoPosted 3 years ago. Direct link to sheeraz.mohammd's post “why does the formula for ...”

why does the formula for the equation of a circle give you centre?

like (x+3)^2+(y+3)^2 = 25

why is the centre -3,3, when plugged in, it clearly does not equal 25•

(2 votes)

Kim Seidel

3 years agoPosted 3 years ago. Direct link to Kim Seidel's post “You are looking at it in ...”

You are looking at it in the wrong way. The formula for a circle is based upon the distance formula for finding the distance between 2 points. In this case one of the points is the center point (-3,3) and you are finding all points that have a distance of 5 units from the center point.

Hope this helps.(8 votes)

Ann Meneses

7 years agoPosted 7 years ago. Direct link to Ann Meneses's post “What if the equation is 9...”

What if the equation is 9x^2 + 24xy + 16y^2 + 90x - 130y = 0 ? I can't figure this out. It says that I should rotate the axes to remove the xy, but I can't really get it.

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(4 votes)

Chris O'Donnell

7 years agoPosted 7 years ago. Direct link to Chris O'Donnell's post “A=9, B=24, C=16 so the di...”

A=9, B=24, C=16 so the discriminant is 576 - 4(9)(16) = 576 - 576 = 0. Therefore this is a parabola. Factorising the first three terms we get (3x+4y)^2 + 90x - 130y = 0. Let u = 3x+4y and v = -4x+3y. Then x = (3u - 4v)/25 and y = (4u + 3v)/25. So the equation becomes:

u^2 + (18/5) (3u-4v) + (26/5) (4u+3v) = u^2 + 158u/5 + 6v/5 = 0

Complete the square: (u+79/5)^2 - 6241/25 + 6v/5 = 0

Rearrange for v: v = 6241/30 - 5/6 (u+79/5)^2

So the directrix is v = 6241/30 + 6/20 = 625/3

And the focus is (u,v) = (-79/5, 6241/30 - 6/20) = (-79/5, 3116/15)Writing in terms of x and y:

The directrix is -12x+9y = 625

And the focus is (x,y) = (-527/15, 112/5)(3 votes)

Bhoke Maseke

3 years agoPosted 3 years ago. Direct link to Bhoke Maseke's post “i still dont get it at t...”

i still dont get it at the practice exercise three

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(2 votes)

David Severin

3 years agoPosted 3 years ago. Direct link to David Severin's post “Exercise 3 requires you t...”

Exercise 3 requires you to complete the square for x and y. So if you have x^2+y^2-10x-16x+53=0, first put variable terms together (x^2-10x+

*__) + (y^2-16y+ ___*) -*___*-*_____*+53=0. To complete the square, you have to divide middle term by 2 and square the result, so -10/2 = -5, (-5)^2=25 and -16/2=-8, (-8_^2=64. Thus, (x^2-10x+25) + (y^2-16x+64)-25-64+53=0. So (x-5)^2 + (y-8)^2 - 36=0.(5 votes)

sugutjune

6 years agoPosted 6 years ago. Direct link to sugutjune's post “How do i expand to get 81...”

How do i expand to get 81 and 49 highlighted? please guide me on the video that might help me more. cos i understand how you did it but i don't understand how to tackle that particular step

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(3 votes)

Cyaniventer

5 years agoPosted 5 years ago. Direct link to Cyaniventer's post “see this video -> https:/...”

see this video -> https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-by-completing-the-square/v/solving-quadratic-equations-by-completing-the-square

sal uses completing the square method to solve it

(2 votes)