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## High school geometry

### Course: High school geometry > Unit 7

Lesson 3: Expanded equation of a circle# Circle equation review

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

## What is the standard equation of a circle?

This is the general $({h},{k})$ with radius ${r}$ .

**standard**equation for the circle centered atCircles can also be given in

**expanded form**, which is simply the result of expanding the binomial squares in the standard form and combining like terms.For example, the equation of the circle centered at $({1},{2})$ with radius ${3}$ is $(x-{1}{)}^{2}+(y-{2}{)}^{2}={{3}}^{2}$ . This is its expanded equation:

*Want to learn more about circle equations? Check out this video.*

## Practice set 1: Using the standard equation of circles

*Want to try more problems like this? Check out this exercise and this exercise.*

## Practice set 2: Writing circle equations

*Want to try more problems like this? Check out this exercise.*

## Practice set 3: Using the expanded equation of circles

To interpret the expanded equation of a circle, we should rewrite it in standard form using the method of "completing the square."

Consider, for example, the process of rewriting the expanded equation ${x}^{2}+{y}^{2}+18x+14y+105=0$ in standard form:

Now we can tell that the center of the circle is $(-9,-7)$ and the radius is $5$ .

*Want to try more problems like this? Check out this exercise and this exercise.*

## Want to join the conversation?

- 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.(25 votes)
- 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(55 votes)

- Parabolas are called Quadratics. What are circle equations called? Are they also called Quadratics because of the square? Thanks.(8 votes)
- 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.(27 votes)

- In problem 2.2 how did you get √17?(6 votes)
- 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).)(22 votes)

- How can you find square roots without a calculator?(4 votes)
- 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)).(8 votes)

- Find the centre and radius of a circle whose equation is x2+y2-4y-21(5 votes)
- 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.(4 votes)

- 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.(2 votes)- 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!(11 votes)

- 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.(4 votes)
- 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)

- 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)- 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.(6 votes)

- i still dont get it at the practice exercise three(2 votes)
- 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)

- 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(3 votes)
- 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)