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### Course: High school geometry>Unit 6

Lesson 3: Problem solving with distance on the coordinate plane

# 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?

• At , why is the Pythagorean Theorem used?
• 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.
• 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.
• 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.
• 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?
• 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.
• 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'.
• 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.
• the lesson questions and the hints have nothing to do this video and its irritating me
• 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?
• Yes, you just need to plug in the x and y values to the equation.
• 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!
• 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.
• 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.
• You can use your method too if you are doing the right thing.