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### Course: Geometry (all content) > Unit 15

Lesson 3: Problem solving with distance on the coordinate plane- Area of trapezoid on the coordinate plane
- Area & perimeter on the coordinate plane
- Points inside/outside/on a circle
- Points inside/outside/on a circle
- Challenge problem: Points on two circles
- Coordinate plane word problem
- Coordinate plane word problems: polygons
- Classifying quadrilaterals on the coordinate plane

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# Points inside/outside/on a circle

Sal uses the distance formula to determine whether the point (-6,-6) is inside, outside, or on the circle centered at (-1,-3) with radius 6.

## Want to join the conversation?

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got it

easy learning(18 votes)- You should write this in the Tips and Thanks section and tell Khan how much you appreciate their work!

Have a BLESSED day!

M.L.M.(5 votes)

- At1:33, why is the Pythagorean Theorem used?(8 votes)
- Try visualizing two horizontal and vertical lines at the x and y values of each of the two points. Now create a line between the points. This creates a right triangle, and you're trying the find the length of the hypotenuse to find the distance between the points.(17 votes)

- Isn't obvious that the point is within the circle? I mean, 6 - 1 is 5, which is less than the radius of 6, and 6-3 is 3, which is less than the radius of 6.(7 votes)
- You mean:

1. Both change in X and change in Y are less than the radius.

2. Hence, the point lies within the circle.

Maybe it's an invalid reasoning.

Counterexample:

- C(0, 0) is the center of a circle that has a radius of 6.

- P(5, 5) lies outside the circle.(8 votes)

- How do you find the radius if you are only given the center (0,0) and a point (-6, √37) that is on the circle?(5 votes)
- Since a radius is a a straight line from the center to the circumference of a circle, you could use the distance formula: √(x2−x1)^2+(y2−y1)^2; to find the distance from the center (0,0) to the point (-6,√37); which would give you the radius.(6 votes)

- If a question says something like:

- Is the point P on the circle? True or False.

Is it safe to interpret the phrase 'on the circle' as 'on the circumference of the circle'?

Or the question sentence is not good and ambiguous?

I think two possible interpretation of the phrase 'on the circle' in the question sentence above.

1. 'on the circumference of the circle'

2. 'on the circumference of the circle or within the circle'.

Interpretation 2 might sound strange but if you think the circle is in 3D world, the circle is just a plane. I don't think it's strange to express 'the point is on the circle' even if the point is within the circumference of the circle.

I know this ambiguity doesn't arise for the problem in the video. I can guess it means 'on the circumference of the circle' because there is another choice says 'Inside the circle'.(3 votes)- Usually, the word 'circle' refers to the circumference of the circle only. If one wants to refer to the circle plus its interior, they'll use the word 'disk.' So the proper interpretation is probably the first one.(7 votes)

- hey so how about if someone gave you the equation for the circle(x^2 + y^2 = r^2)? can you just plug in the coordinates for p and check if that it simplifies and equals the radius squared or not?(3 votes)
- Yes, you just need to plug in the x and y values to the equation.(3 votes)

- Why can't I just use the distance formula rather than the Pythagorean theorem when we try to find out the length of the center of the circle to the point? I am very annoyed by this and cant get it right ever.(3 votes)
- You can use your method too if you are doing the right thing.(3 votes)

- Why does sal need to do the square root for the formula? Is it nesessary for him to do so? And if so can u please explain why?(2 votes)
- That is part of the distance formula which is an application of the Pythagorean Theorem. Since c^2 is in the formula, the opposite of squaring something is the square root.(5 votes)

- the lesson questions and the hints have nothing to do this video and its irritating me(4 votes)
- I was working through a problem when I came across something I didn't understand. I'd love it if someone could help me answer this or prove one step wrong to explain it! Here's what happened:

2√3

I thought "two square roots of three / two x square root of 3". Well 2 x √3 should be the same as √3 + √3. I wonder if this is what I got wrong somehow but let's continue.

= √3 + √3

= √3 + 3 --- combining the numbers under the radical

= √6

^

2 x 3

So that's all it can be simplified.

But, 2√3 can be read as the result of already simplifying a square root. So let's do the simplification backwards.

2√3

= √2 x √2 x √3

= √2 x 2 x 3

= √12

So that's my problem: √6 ≠ √12. I know I probably did something wrong, and that it's probably super obvious but I just can't tell. Please help me out here.

Thanks!(3 votes)- You cannot add the numbers under a radical. √3 +√3 ≠√(3+3), and your calculation here proves this.

To take the square root of something is to raise it to the 1/2 power. Exponents distribute over multiplication (which is why you can say √2 ·√2 ·√3=√(2·2·3)), but not over addition.(2 votes)

## Video transcript

- [Voiceover] A circle is
centered at the point C, which has the coordinates negative one, comma, negative three. And has a radius of six. Where does the point P, which has the coordinates negative six,
comma, negative six, lie? We have three options. Inside the circle, on the
circle, or outside the circle. And the key realization here is just what a circle is all about. If we have the point C, which
is the center of a circle, a circle of radius six, so
let me draw that radius. So we say that is its radius. Is six units. The circle will look something like this. Remember, the circle is
a set of all points that are exactly six units
away from that center. So that's the definition of a circle, it's a set of all points that are exactly six units away from the center. So if, for example, P is
less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle, and if it's more than six units away, it's going to be outside of the circle. So the key is, is let's
find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we
are outside of the circle. So let's do that. So if we wanted to find,
well there's different notations for the distance. Well, I'll just write D, or I could write the distance between C and P is going to be equal to. And the distance formula
comes straight out of the Pythagorean Theorem. But it's going to be the square root of our change in X squared plus
our change in Y squared. So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, and we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one. And we're going to square it. So what we have inside
here, that is change in X. So we're taking our change in X squared, and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative
six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in
Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative
six plus positive one, is one way to think about it, so this is negative five squared. And then this is negative six plus three. So plus negative three squared. And once again, you can see, our change in X is negative five. We go five lower in X, and
we're going three lower in Y. So we're changing Y, is negative three. So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus
nine, which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater
than six, or equal to six? Well, we know that six is equal to the square root of 36,
so the square root of 34 is less than the square root of 36, so I could write the square root of 34 is less than the square root of 36, and so the square root
of 34 is less than six. These are the square root of 36, is six. And so since the distance between C and P is less than six, we are going to be on the inside of the circle. If I somehow got square root of 36 here, then we'd be on the circle, and if I somehow got
square root of 37 here, or something larger, we would
have been outside the circle.