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### Course: Get ready for Precalculus>Unit 7

Lesson 2: Standard equation of a circle

# Features of a circle from its standard equation

Sal finds the center and the radius of the circle whose equation is (x+3)^2+(y-4)^2=49. Created by Sal Khan.

## Want to join the conversation?

• I'm stuck, this video was very helpful, but i have an equation (X^2+Y^2=144) and i cant find the center. I feel like it is one of those things where when you are told, you do a face palm. Thanks.
• The center is at the origin.
The equation of a circle in standard form is:
`(x−h)²+(y−k)²=r²`
Where `r` is the radius and (h,k) is center.
If either `-h` or `-k` is missing, then its value must be 0. Thus, if both are missing the circle must be centered at the origin, (0,0).
• I understand how Sal derives the formula, given the diagram with the generic variables, but what I do not understand is why this equation actually plots a circle.

Any help would be appreciated.
• A circle is the set of all points that are equidistant from a central point. (h,k) gives is the centre of the circle and the distance gives us the radius we can then create our circle.
• At the end why did you say h is -3 when it was 3 can you help me out cause that's confusing to me why it would change to a different value
• It's because of the negatives in front of the 3. The equation has
(x-h)^2. See how it's "minus h"? That's the original equation. The example problem had a similar version to the equation, except it said
"(x+3)^2". This is the final version of the original "(x-h)^2". Sal was trying to work backwards and try to find what the equation was when there was a "minus" sign instead of a plus. We know that two negatives (in this case, "x - -3", or "x minus negative 3") will give a positive answer ("x + 3"). That's why the -3 became positive. The actual value of "h" is -3, because "h" has to be subtracted from the x-value. When h is equal to -3, the equation would be "(x-h)^2", or "(x- -3)^2", which then simplifies to
"(x+3)^2".

Hope this helps clear things up to whoever has the same question! If you still don't understand, I think it's a good idea to re-learn/practice double negatives :)
• why is it x-h, ans y-k, and not k-y or h-x, what's the difference?
• I am sorry but I think answer given by Maithew isn't correct at all.
I'll tell you the reason why.
h and k are some center point of the circle. Let's assume it as 3 for h and 4 for k.
Let's take R to be 2.
Let's assume(I hope you understand because I cannot draw it) we take a point which is at extreme upper right(north east) of the circle and take those point as x and y.
Now to get the distance to that point from center of circle we cannot simply take x and y because it includes the distance (3,4) of h and k. Therefore to get the exact distance we have to subtract h and k from x and y.
I hope it helps and doesn't confuse you more.
• related to this i got a question .if in a circle the two chord are parallel each with lenth 24 and 10cm .they is a distance of 17 cm betwenn the two chords.then what is the radius of the circle
• The end point of one cord and the end point of the other cord are on the circle. You can't draw a line to the center of the circle because the center isn't given. A perpendicular bisector will give you a 90° at the two cords. Now we have right triangles that have congruent hypotenuses. One triangle will have a 5 cm leg and the other a 12 cm leg. Now, for the hypotenuses to be congruent the legs of the triangles must be congruent, therefore both triangles have 5 and 12 as legs. The only math you need is 5² + 12² = r².
• Hey, what is an ellipse? I keep seeing it in Geometry and I don't know what it means...thank you!
• An ellipse is an oval. If you slice the tip of a cone off (at a shallow enough angle that the cut doesn't hit the bottom of the cone), the new edge of the cone will be an ellipse.

An ellipse has two diameters, the longest line you can draw through the center, and the shortest. If the two diameters happen to be the same length, you get a circle. So a circle is a special type of ellipse.
• What if you have x^2 + (y-2)^2=16 ? I know the radius would be 4, and the Y coordinate is -2. But what would the X coordinate be? Would it be zero?
• Well lets try expanding it out. I can subtract zero from anything and not change it's value so
(x-0)^2 + (y-2)^2 = 16. Hope that makes it clear.
• Can you use the Pythagorean theorem to replace the distance formula?
• The distance formula is a variation of the Pythagorean Theorem. The only differences are:
1) In the distance formula you need to calculate the distance between 2 points: (x2-x1) and (y2-y1). In the Pythagorean theorem, these are known.
2) The distance formula is solved for "d" (or the hypotenuse) where the Pythagorean theorem is not yet solved for the hypotenuse.
• At about 3 minutes into the video Sal wrote x - - 3 ^2, shouldn't it be x - + 3 ^2?
• equation: (x+3)^2 + (y-4)^2 = 49

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

now compare the equation given in question against standard equation, some may say they can't because the original has (x+3) but the standard one has (x-h) so to explain that sal made it (x-(-3)).

you're saying it should've been (x-(+3)) but that's just wrong. why? -> (x+3) is not equal to (x-(+3)) you are chaning the question.

whereas sal wrote (x-(-3)) which is same as (x+3), think about this again. you can also substitute values for x to make it easier. (i.e. take x=0 & substitute in both and see which is correct)
• This would be much clearer if the circle were plotted in a coordinate plane to ease appreciation of subtracting h from x and k from y. Thanks!

## Video transcript

The equation of a circle C is x plus 3 squared plus y minus 4 squared is equal to 49. What are its center h, k and its radius r? So let's just remind ourselves what a circle is. You have some point, let's call that h, k. The circle is the set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from h, k. So let's say that this distance right over here is r, and so we want all of the set of points that are exactly r away. So all the points x comma y that are exactly r way. And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it, but you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and h, k, and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. In fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here, that essentially is the change in the vertical axis between these two points, up here, we're at y, here we're k, so this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle, because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here, so these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared, and this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set-- this describes any x and y that satisfies this equation will sit on this circle. Now, with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote, we just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus-- well this is already in the form-- plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3-- I want to do that in the red color-- that our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. Make sure to get-- you know you might say, hey, there's a negative 4 here, no. But look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3, but no you're subtracting the h. So you'd say minus negative 3, and similarly, the radius is 7.