Main content

### Course: High school geometry > Unit 6

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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Challenge problem: Points on two circles

Watch Sal solve a challenging problem where he has to determine if points are on both, one, or neither of two circles. Created by Sal Khan.

## Want to join the conversation?

- Would D not be considered to be "In' the circle, why was it classified as "Neither?"(16 votes)
- The question was whether the point is on circle A, circle B, or neither. We don't care if it's inside or outside of either circle.(16 votes)

- At2:12,It is mentioned how we find the (Ra)2.How do we do it i don't get the (-5-0)2(7 votes)
- it is straight up the Pythagorean theorem we can derived a formula which is what we call the distance formula: d=sqrt[(x_2-x_1)^2 + (y_2-y_1)^2] (*We used distance formula iff we're given two points to identify the displacement of those points) . In sal example, he wants to find the radius between center A and point P after using the Pythagorean theorem (or distance formula) he conclude that the radius is 5*sqrt(2).(5 votes)

- Why not draw the 2 circle and place the points and see if these are on circle A and B ?(8 votes)
- Sure, a graphical solution is one way of solving the problem - and in this case, it might actually be easier. When you're not dealing with integers, though - for example, if you want to know if the point P(e, pi) lies on some circle - you're going to have to do that using the distance formula, so it's best to learn the method.

He kind of does that in this video for points D and E, but it's probably best to do it properly with compasses and grid paper.(10 votes)

- Wait, I understand all the math that he used to get to the solutions, however, when it says "Point P is ON circles A and B" how do we know that it is on the perimeter of the circle? When a point is on a circle, can't it be anywhere on the circle, therefore deeming this problem unsolvable??(4 votes)
- The definition of a circle in analytic geometry is: the set of all points located the same distance away from a given point and is algebraically defined by the equation
`(x-h)^2 + (y-k)^2 = r^2`

where (h,k) is the center point and r is the radius. There is no magic equation, as far as I know, that defines all points on the circumference and inside a circle, so in your interpretation of a circle there is no*equation*to solve that problem. You could, however, turn the above equation into an inequality to determine if "Point P"`(x,y)`

is a certain distance`(r)`

from the center of the circle`(h,k)`

.`ex. if: (x-h)^2 + (y-k)^2 <= r^2`

then it is on the circle(12 votes)

- What's a locus? I've read the dictionary definition but what does it encompass? What can you use it to describe?(5 votes)
- looks like...'locus' means all the answers to a kind of problem that has a group of solutions instead of just one exact one.

the word is latin-ish for "this is where you'll find it"(5 votes)

- 5:19....Is Sal assuming that it is a right angle?(7 votes)
- No. It's a coordinate plane(1 vote)

- While solving point C, why is the distance formula only applied with reference to point B and not A or P?(3 votes)
- Sal took the sqrt8 and turned it into 2*sqrt2 ,so why didn't he take sqrt10 and simplify it to 5*sqrt2?(1 vote)
- √8=√(4•2)=√4•√2=2√2

√10=√(5•2)=√5•√2. There are no perfect squares to work with in the factorization of 10.(5 votes)

- In the video he talks about eyeballing points and not dealing with points that are obviously not on the circle but if you dont have a graph do you have to calculate all that?(2 votes)
- I know this is a bit late, but...
**IF**you don't have a graph, you can still eyeball the points based off the circle. You can look to see whether the point is outside the circle or inside. However, in cases where questions do**not**provide a circle or a graph but only coordinates or equations, the best thing to do is to solve it.

Hope this helps!(3 votes)

- How do you know that the circles don't overlap and that point P (0,0) is in both of the circles? Would you know because the question says that point P is on circles A and B and doesn't say that point P is in circles A and B?(2 votes)
- Yes, the problem says it is ON circle A/B, if it said inside, this problem couldn't be solved.(2 votes)

## Video transcript

Point A is at
negative 5 comma 5. So this is negative
5 right over here. This is 1, 2, 3, 4, 5. That's 5 right over there. So point A is right about there. So that is point A, just like
that, at negative 5 comma 5. And then, it's a center
of circle A, which I won't draw just yet, because I
don't know the radius of circle A. Point B is at--
let me underline these in the appropriate color--
point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B
right over there, so center of circle
B. Point P is at 0, 0, so it's right over
there at the origin. And it is on circles A and B. Well, that's a big
piece of information. Because that tells us, if
this is on both circles, then that means that
this is B's radius away from the point
B from the center. And this tells us that it
is circle A's radius away from its center,
which is at point A. So let's figure out what
those radii actually are. And so we can
imagine-- let me draw a radius or the
radius for circle A-- we now know
since P sits on it, that this could be considered
the radius for circle A. And you could use a distance
formula, but what we'll see is that the distance
formula is really just falling out of the
Pythagorean theorem. So the distance formula tells
us the radius right over here, this is just the distance
between those two points. So the radius or the distance
between those two points squared is going to be equal
to our change in x values between A and P. So
our change in x values, we could write it as
negative 5 minus 0 squared. That's our change in x,
negative 5 minus 0 squared, plus our change in
y, 5 minus 0 squared, which gets us that our distance
between these two points, which is the length
of the radius squared, is equal to negative 5
squared plus 5 squared. Or we could say
that the radius is equal to the square root of,
this is 25, this is 25, 50. 50, we can write as 25 times 2. So this is equal
to the square root of 25 times the square
root of 2, which is 5 times the square root of 2. So this distance right over
here is 5 times the square root of 2. Now, I said this is just the
same thing as the Pythagorean theorem. Why? Well, if we were to construct a
right triangle right over here, then we can look
at this distance. This distance would be the
absolute value of negative 5 minus 0. Or you could say, this
is 0 minus negative 5. This distance right
over here is 5. This distance is the
distance between 0 and 5 in the y direction. That's 5. Pythagorean theorem tells
us that 5 squared, which is 25, plus 5
squared, another 25, is going to be equal to
your hypotenuse squared. And that's exactly
what we have here. Now, you might be
saying, wait, wait, wait, this thing had a
negative 5 squared here, while here, you
had a positive 5. But the reason why we could
do this is when you square it, the negative disappears. The distance formula,
you could write it this way, where you're
taking the absolute value. And then, it becomes very
clear that this really is just the Pythagorean theorem. This would be 5
squared plus 5 squared. 5 squared plus 5 squared. The reason why you
don't have to do this is because a sign doesn't
matter when you square it. It always will become
a positive value. But either way, we
figured out this radius. Now, let's figure out the radius
of circle B. Same exact idea. The radius of circle B squared
is equal to our change in x. So we could write as 3
minus 0, or 0 minus 3. But we'll just write
it as 3 minus 0. 3 minus 0 squared plus
1 minus 0 squared. Or the radius or the distance
between these two points is equal to the square
root of-- let's see, this is 3 squared
plus 1 squared. That's 9 plus 1. This is the square root of 10. The radius of B is
the square root of 10. Now, they ask us, which
of the following points are on circle A, circle
B, or both circles? So all we have to do now
is look at these points. If this point is the square
root of 10 away from point B, then it's on the circle. It's a radius away. A circle is the locus
of all points that are a radius away
from the center. If it's 5 square roots
of 2 from this point, then it's on circle A. If it's
neither, then it's neither. Or it could be both. So let's try these
out, one by one. So point C is at 4, negative 2. So let me color
this in a new color. So point C-- let me do
it in orange-- point C is at 1, 2, 3, 4, negative 2. Point C is right over there. Now, it looks pretty close. Just, this is a hand-drawn
drawing, so it's not perfect. So point C is there. It looks pretty close. But let's actually verify it. The distance between
point C and point D, so the distance squared, is
going to be equal to the change in x's. So we could say, 4 minus--
so we're trying [INAUDIBLE] C and B, is 4 minus 3 squared
plus negative 2 minus 1 squared, which is equal
to-- This is 1 squared plus negative 3 squared. And so our distance
squared is equal to 10. Or our distance is equal
to the square root of 10. So this is also, the
distance right over here is the square root of 10. So this is on the circle. If we wanted to
draw circle B, it would look something like this. And once again, I'm hand-drawing
it, so it's not perfect. But it would look
something-- I'm going to draw part of it--
it looks something like this. This is exactly a radius away. So let me write, this
is on circle B. Now, let's look at this point. The point 5 comma 3. So I'll do that in pink. So 1, 2, 3, 4, 5 comma 3. So this looks close, but
let's verify, just in case. So now, our distance is equal
to-- let me just write it this way-- our
distance squared is going to be our
change in x squared. So 5 minus 3 squared plus 3
minus 1 squared, change in y. And so our distance is going
to be equal to-- actually I don't want to
skip too many steps. Let's see, this is
2 squared, which is 4, plus 2 squared,
which is another 4. So our distance is going to be
equal to the square root of 8, which is the same thing as
the square root of 2 times 4, which is the same thing as 2
times the square root of 2. Square root of 4 is 2. And then, of course, you just
have the 2 left in the radical. So this is a different distance
away than square root of 10. So this one right over here
is definitely not on circle B. And just eyeballing
it, you can see that it's not going to be on
circle A. This distance, just eyeballing it, is much further
than 5 square roots of 2. And that's also true
for point C. Point C is much further than
5 square roots of 2. You can just look
at that visually. They're much further than
a radius away from A. So this point right over
here, this is neither. This is on neither circle. Now, finally, we have the
point, negative 2 comma 8. So let me find-- I'm
running out of colors. Let me see, I could use, so I
guess I'll use yellow again. Negative 2 comma 8, so that's
negative 2 comma 1, 2, 3, 4, 5, 6, 7, 8. So it's right over here. That is point E.
Just eyeballing it, this distance-- so it's
clearly way too far. Just looking at,
just eyeballing it, it's clearly more than
a radius away from B. So this isn't going
to be on circle B. And also, looking at
it relative to point A, it looks much closer to point
A. It doesn't even seem close, than point P is. So it looks, just inspecting
it, that you could rule this one out, that this is
going to be neither. But we can verify this
on our own, if we like. We can just find the distance
between these two points. Our distance squared is going
to be our change in x's. So negative 2 minus negative 5
squared plus our change in y. So it's 8 minus 5 squared. And so this is, our
distance squared is going to be equal to
negative 2 minus negative 5. That's negative 2 plus 5. So that's going to be 3
squared plus 3 squared. And you see that right over
here, Pythagorean theorem. This distance right
over here is 3. This distance right over here,
this is your change in x is 3. Change in y is 3. 3 squared plus 3
squared is going to be the distance squared,
the hypotenuse squared. So our distance squared is
going to be-- or I could say, our distance--
skip a few steps-- is equal to the square root of--
we can write this as 9 times 2 or the distance is equal to
3 times the square root of 2. The radius of circle A is 5
times the square root of 2, not 3 times the
square root of 2. So this is actually going
to be inside the circle. So if we want to
draw circle A, it's going to look
something like this. And point E is on the inside. Point D and point C are on
the outside of circle A. The only one that sits on any
of the circles is point C.