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## High school geometry

### Course: High school geometry > Unit 7

Lesson 2: Standard equation of a circle- Features of a circle from its standard equation
- Features of a circle from its standard equation
- Graphing a circle from its standard equation
- Graph a circle from its standard equation
- Writing standard equation of a circle
- Write standard equation of a circle

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# Graphing a circle from its standard equation

Sal graphs the circle whose equation is (x+5)²+(y-5)²=4.

## Want to join the conversation?

- this corona stuff is hard becuase i have to do online school and its hard but im working my hardest can i get a shoutout.(18 votes)
- Keep up the good work! Hope that colleges/schools open again and stay open!(4 votes)

- at1:52,why did (x+5)^2 became (x- -5)^2 ? I'm doomedddd(5 votes)
- Sal wrote it that way to get it into the form of the equation for a circle, which is

(x - h)^2 + (y - k)^2 = r^2. In other words, the center of the circle is at (h,k), where h and k are the numbers being SUBTRACTED from x and y.

Hope this saves you from being doomedddd!(15 votes)

- this is not so clear to me yet so can anybody answer? how do you estimate how wide the circle is? i get that the radius is like the distance from the circle? so like in the first part of the video it is 4 points away from the center??(2 votes)
- The 4 in the video is the radius squared. You need to take the sqrt(4) = 2

This tells you the distance from the center of the circle to the edge (the radius) = 2.

Then, 2 * radius = the diameter of the circle (the total width). In this case it also happens to be 4.

Lets say the equation is: x^2 + y^2 = 25

From the equation we can tell:

1) The center of the circle is at (0, 0)

2) Sqrt(25) = 5. So, the radius of the circle = 5 (center of circle to the edge = 5). Thus, the diameter of the circle = 2*5 = 10 (entire width of the circle = 10)

Hope this helps.(16 votes)

- What makes the x plus 5 turn into x minus negative 5?(4 votes)
- The standard equation of a circle is (x-h)^2+(y-k)^2=r^2, where (h,k) is the center. The equation Sal started with was not in this format. Thus, in order to be able to graph the circle, he needed to change the
*appearance*of the equation. We don't change the equation itself, though; we don't change any values.

Remember that a double negative equals a positive. In other words, two subtraction signs equals an addition sign. So that's what Sal did here, in order to put the equation in standard format. He merely changed how we see it.

Hope this helps!(4 votes)

- for the first one, how do we know it's negative five and not five? and same for the y, how do we know if it's positive or negative?(2 votes)
- The form of the equation is (x = h)^2 + (y -k)^2 = r^2, so you always have to change the sign of what is inside to find the point. This may not help, but I think of it as wanting to have the inside equal to 0 to see how much it moves, so if x - h = 0, then x = h.(5 votes)

- What will be the type of the triangle if 2 edges other than centre touches to the circle?Acute or Obtuse or Right angle triangle?(3 votes)
- If you always use the centre as one of the points, only 1 edge can touch the circle. After all, the other two edges must leave the circle to end up at the centre.

If all points are on the circle, then you can have any type of triangle. In fact, you can pick any random triangle at all, and you'll always be able to draw a circle around it such that the triangle's corner points (called "vertices") will all be on the circle.(3 votes)

- whats a center?(2 votes)
- The center of a circle is the point in the middle of the circle. All points that create the circle are an equal distance from that central point.(4 votes)

- so the equation of the circle can nor be represented by function?because for each x value we have two y valuse?(3 votes)
- Yes.you are right...

A function is a relation in which one variable specifies a single value of another variable. For example, when you toss a ball, each second that passes has one and only one ... We know that y is a function of x because for each x-coordinate there is ... can have more than one output (y-coordinate), and y is not a function of x(1 vote)

- How do I find the center and radius of the equation

(x+2 radical sign 3)squared +(y +2 radical sign 3) squared = 1(2 votes)- The equation of a circle is (x-h)^2+(y -k)^2= r^2, with the center being (h,k) and radius r. That means that the center of this circle is (-2 radical sign 3, -2 radical sign 3) and the radius is 1 (because 1^2=1)(2 votes)

- So, can I basically subtract either the
`(h,k)`

from the`(x,y)`

or the`(x,y)`

from the`(h,k)`

*as long as I'm consistent throughout the problem*, since I will be squaring the answers?(1 vote)- For your calculations, it doesn't matter. But it's important to remember that the standard formula is given with (x-h) and (y-k), so if you are asked for the value of h (or if you wish to communicate it to someone else), it's best to follow the standard order.(3 votes)

## Video transcript

- [Voiceover] Whereas to graph the circle x plus five squared plus y
minus 5 squared equals four. I know what you're thinking. What's all of this silliness
on the right hand side? This is actually just the view we use when we're trying to debug
things on Khan Academy. But we can still do the exercise. So it says drag the center point
and perimeter of the circle to graph the equation. So the first thing we
want to think about is well what's the center of this equation? Well the standard form of a circle is x minus the x coordinate
of the center squared, plus y minus the y coordinate
of the center squared is equal to the radius squared. So x minus the x coordinate of the center. So the x coordinate of the center must be negative five. Cause the way we can get
a positive five here's by subtracting a negative five. So the x coordinate must be negative five and the y coordinate
must be positive five. Cause y minus the y
coordinate of the center. So y coordinate is positive five and then the radius squared
is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, we could drag this out like this, but this the way it's drawn, the radius is indeed equal to two. And so we're done. And I really want to hit the point home of what I just did. So let me get my little scratch pad out. Sorry for knocking the
microphone just now. That equation was x plus five squared plus y minus five squared is equal to four squared. So I want to rewrite this as, this is x minus negative five, x minus negative five squared, plus y minus positive five, positive five squared is equal to, instead of writing it as four I'll write it as two squared. So this right over here tells us that the center of the
circle is going to be x equals negative five y equals five and the radius is going
to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of
the Pythagorean Theorem, straight out of the distance formula, which comes out of the
Pythagorean Theorem. Remember, if you have some center, in this case is the point
negative five comma five, so negative five comma five, and you want to find
all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them, x comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a
circle with radius two around that center. Plus think about how we
got that actual formula. Well the distance between that coordinate, between any of these x's and y's, it could be an x and y here,
it could be an x and y here, and this is going to be two. So we can have our change in x. So we have x minus negative five. So that's our change
in x between any point x comma y, negative five comma five. So our change in x squared plus our change in y squared, so it's going to be y minus the y coordinate over here, squared is going to be equal to the radius squared. So the change in y is going to be from this y to that y, this is the end point. The end minus the beginning y minus five, y minus five squared. And so this shows for
any x y that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write
that, is equal to four. And let me make it, I
really want to, you know, I dislike it when formulas
are just memorized. You don't see the
connection to other things. Notice we can construct a nice little right triangle here. So our change in x is
that right over there. So that is our change in x. Change in x. And our change in y our change in y, not
the change in y squared, but our change in y, is that right over there. Change in y, our change in y you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five. x minus negative five. So this is just change in x squared plus change in y squared is equal to the hypotenuse squared, which is the length of,
which is this radius. So once again, comes straight out of the Pythagorean Theorem. Hopefully that makes sense.