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## End of the String

Now, we're going to move away from the distance formula and the Pythagorean theorem. We'll still use the Pythagorean theorem in this upcoming lesson, but we'll use it for circles. We're almost to the end of the course, but before we get to that end, let's talk about the end of a string. Here's a string of unknown length. Now, I'm not going to tell you what length it is yet. You're actually going to calculate that yourself, we're going to take this string and place it on a coordinate grid. Notice that I have one end of the string at the origin 0, 0. The horizontal distance from the center which is the origin to the end of the string is 40 centimeters, that's this distance. And the vertical distance from the origin to the end of the string is 30 centimeters. So, knowing this horizontal distance, and this vertical distance, what x, y coordinate can describe the position of the end of this string? Write that answer here. Once you've done that, I want you to try and figure out the length of the string. How long is this? Write that answer here.

## End of the String

The end of this string is at the point 40, 30, and the link to this string is the string we just want to have a coordinate pair. The x distance is 40 units or 40 centimeter, and the y, or vertical distance is 30, 30 centimeter. The length of this string is actually the hypotenuse of this right triangle. So we can use the Pythagorean theorem to solve for the length of the string. We'll have a squared, which is 30, plus b squared, which is 40, equal to c squared, our hypotenuse. When we square these two numbers we'll get 900 plus 1600. Then when we sum these numbers we will get 2500 equal to c squared. Now we just want to take the square root of both side of our equation to solve for c. Taking the square root of both sides, will only keep the positive root, since we're dealing with a distance, so c equals 50 centimeter. Now the much faster approach, to solving this, would be to understand that we have a 3, 4, 5 right triangle. The ratio has just been multiplied by 10. So, this side is 30, this side is 40. And this length of our string, is 50 centimeters. If you solved it using that 3, 4, 5 ratio, amazing work.

## Equal Sides

Let's draw another red triangle Here is my string and the end of it is at the origin. Here is a horizontal distance and here is a vertical distance. Now between these two triangles that we created which two side lengths are the same? You can check which of the sides you think are equal.

## Equal Sides

Well we know that these two hypotenuses must be the same. This hypotenuse, and this hypotenuse were both measured using a length of the string. The length of the string is 50cm and it hasn't changed since when we created this triangle from the first triangle. We only change the position of the end of the string when creating a second triangle. So the length of the string didn't change so we know these two lengths must be equal.

## Another Triangle

So we have this new right triangle but what is this vertical distance here? When you think you know it, type it in here and I have already included the units of centimeters. You will also want to round your answer to the nearest tenth. So you should have one digit after the decimal point.

## Another Triangle

This vertical distance is about forty five and eight tenths of a centimeter. Nice problem solving if you got this correct. We'll find this vertical leg by using the Pythagorean theorem. We know the horizontal distance are one leg of the triangle is twenty units. And we know the length of a string is fifty units. Now our units are centimeters but I'm just leaving it out as we work through the problem. We want to make sure we add that back in when we get to our answer. So twenty squared equals four hundred and fifty squared equals two thousand five hundred. Then we'll subtract four hundred from both sides of our equation to get b squared equals two thousand one hundred. We'll take the square root of both sides of the equation indicating plus or minus. But we'll only use the positive root in the sense that we have a distance. In this case b is about forty five and eighty two hundredths Round it to the nearest tenth we'll have forty five and eight tenths. We've been working in centimeters so we'll report this answer in centimeters.

## Another Position

Let's make some more of these right triangles in this first quadrant. Here is another red triangle and here is another one as well. So, notice I have four different red triangles in the first quadrant and they are all related by the Pythagorean theorem. The length of the string is fifty centimeters in each case. So that means the hypotenuses of each right triangle is fifty centimeters. But let's create a right triangle in this quadrant. If I had this right triangle what would be the coordinates of the end of the string now? Write the coordinates for this position here.

## Another Position

This coordinate would be negative 40, 30. Nice thinking if you got these two numbers. Even though the distance is reported as a positive value we need a negative x coordinate in order to indicate a different direction. If we just had 40, 30 then the end of the string would be the same position that we described first over here. We moved in the negative x direction so we use negative 40 for the x coordinate and then positive 30 for the y coordinate.

## Lots of Right Triangles

So, let's say we created lots of right triangles. What if I made another triangle here and another triangle here and another one here? And we could keep going and make another right triangle here and another one right here. Now, the points along the outer edge represent the end of the string when making any one right triangle. What shape would all these points, that represent the position of the end of the string, make?

## Lots of Right Triangles

These points would form a circle. I hope you knew that was coming. Great job if you got it. When we think about a circle, we're really thinking about a particular relationship between any of the points. That is, the distance from any one of these points to the origin is always the same. In this case, any distance from the end point of a string to the origin represent 50 centimeters. They're all constant.

## Equation for the Position of the End of the String

So let's lose our coordinate grid and all of those right triangles. That's right, we've gone off the grid. We just have one piece of string centered at the origin, 0,0. And we have the end of the string at some point x comma y. We know that if we were to rotate this piece of string anywhere, we would create a circle by the position of the end of the string. So, what's that equation that we can write relating x and y, and the length of the string? If you think about relating the horizontal distance and this vertical distance with the length of the string, I'm sure you can come up with the correct equation.

## Equation for the Position of the End of the String

The equation for our circle is x squared plus y squared equals 2500. Now that's amazing if you got this equation right. This one wasn't easy. Now, we know this horizontal distance is x units. That's the distance between these two points, at least just the horizontal distance. And we know that this vertical distance is a distance of y units. We started at 0 and we went down y. So, notice that we've created a right triangle. So we can relate these distances with this side length. The string is 50 centimeters in length, so we know that x squared plus y squared equals 50 squared, or 2500. So, no matter where the position of the string is, this distance relationship will always be true. X squared plus y squared will always equal 2,500. And it turns out that this distance from the center to any point on the circle is called the radius. In this case 50 centimeters.

## Graph of a Circle

So, we know that this equation describes this circle. The distance from the center of the circle to any point on the circle will always be 50 centimeters. And that should make sense, because that's the length of our string. No matter where the end of the string is, we know that the distance from the origin to the circle is 50 centimeters. Now, this distance is always constant and it's called the radius of the circle. So in this case, the length of the string is the radius and it's 50 centimeters.

## Equation of a Circle Centered at the Origin

So now let consider any equation of a circle that's centered at the origin. So, here's a new piece of string. And I'm going to cut it. So that way, I don't know the radius. So we'll let the length of this string be r. We don't know it. And at this length, we'll represent the radius of our new circle. And remember, we can create this circle by plotting all the points of the end of the string. If we do this, what equation describes this circle? Write that equation here. Now, if you're a little stumped. Remember that you're centered at the origin. And any point out here is x, y.

## Equation of a Circle Centered at the Origin

The equation would be x squared plus y squared equals r squared. Now that's amazing work if you got this equation. Remember that for any point on our circle, we can set up a right triangle. So, we'll have this leg squared, plus this leg squared, equals our radius squared. It works for a point out here. And it would've worked for this point, down here. We have another right triangle, here, so we would have X squared plus Y squared equals r squared. Where r is the length of our string.

Now that you've seen the general equation for circle centered at the origin, what do you think would be the radius of this equation? Keep in mind that the equation really represents a circle.

This equation would have a radius of 8. Great thinking if you found that number. We know that the general equation for a circle centered at the origin is x squared, plus y squared, equals r squared. So, we know that r squared must equal the number 64. Its the number on the end of the equation. So, solving for r will take the square root of both side so, one r equals positive 8, and I use the positive root only since we are dealing with radius. A distance from the center to the points on the circle. If we actually graph the circle, we could easily see that the center is at the origin. And that the radius is 8, the distance to any point on the circle.

## Equation of a Circle Check

Now, we haven't created any sort of equations yet, but I think you'll be able to answer this question. What would be the equation of a circle that has a radius of 10 and is centered at the origin. Your equation should use the variables x and y and when you think you have it enter it here

## Equation of a Circle Check

The equation would be x squared plus y squared equals 100. Great work if you got that equation. We know, this is the equation of any circle centered at the origin. We just need to know the radius, r. In this case, the r value is 10. So, we just plug in 10 for r. We'll have this equation. And then simplifying 10 squared, we'll get 100. So, this is the equation for our circle.

## Points Inside Outside On a Circle

We just saw that this equation was a circle, with a radius of 10. So now let's consider some other points. Do you think each of these points would be, inside of our circle? Outside of our circle? Or on the circle. Choose one of these for each of the points. Keep in mind if it's on the circle, this point, the x and y value, would satisfy our equation. If it doesn't satisfy the equation, then think about what kind of numbers our results would be inside the circle and what kind would be outside. Good luck here.

## Points Inside Outside On a Circle

These two points are inside the circle, this point is on the circle and this point is outside the circle. Great work if you got all four of them correct. Now if you weren't sure where to start, let's look at this first point and see why it's on the circle. Then you might be able to figure out why these are inside and this one's outside. Remember that this is really just a right triangle in space. It has some horizontal distance x and some vertical distance y. So we want to be sure that when we plug in the values for x like negative 6 and the value of y like positive 8, we get 100. So squaring negative 6 and squaring 8 we'll get 36 plus 64 which equals 100. So yes, it turns out that this statement is true. This means that the point negative 6, 8 Is a point on our circle, since it satisfies the equation. We also know that this point is on our circle since we can set up a right triangle. The point, negative 6, 8 is 6 units left from the origin and 8 units up. If we use the Pythagorean theorem here, we'll discover that this side, or this radius, is 10. This should also make sense because we really have a 3, 4, 5 triangle. The triangle has just been multiplied by 2 to give us a 6, 8, 10 triangle. So we know that the distance from the origin to this point is 10 units. And that matches with our idea that any point on our circle is 10 units from the center. So, by checking the point in our calculations or by drawing out a diagram, we can know that this point is definitely on our circle. For this point, it turns out it will be inside. When we check the point negative 9, 4, we'll plug in x with the value of negative 9, and we'll substitute the value of y with 4. Simple, find the left-hand side of our equation we'll get 97. We know that's not equal to 100, and in fact this is less than 100. Since negative 9 squared plus 4 squared equals 97, we really know that we're inside the circle. Remember that 10 is the radius of our circle, it's the square root of this number. So, this means the distance from the origin to this point is the square root of 97 which is definitely less than 10. The right triangle for this point would look like this we'd have 9 units to the left of the origin and then 4 units up. This point would just be shy of reaching our circle. We know that the distance from the origin to this point is the square root of 97. It's the square root of this number. We really want this distance to be 10 or the square root of a 100 and its not, its less than that. So that's how we know that this point is inside the circle. The same would be true for the point 7, negative 7. If we plug in x with the value of 7 and y with a value of negative 7 we'll get the statement 98 equals 100. This isn't true so we know that the distance from the origin to this point is less than 10. So this point must be inside. And finally for this point when we plug in x with a value of 5. And y with the value of 9, will get that means that the distance of 5, 9, from the origin, is greater than 10. The point, 5, 9 would be, just outside of our circle. We can even see this one desmos. Here's the equation of our circle and we can see that the point negative 6, 8 is on it. Even if we zoom in, we can quickly see that, yes, the point is on the circle. And then this other point negative 9, 4 isn't. It's just slightly inside of our circle. The other point that we said was inside the circle is 7, negative 7. This point down here. Zooming in on this region, we see that it's just shy of being on the circle. And that should make sense, because we found that the distance from the origin to this point, was the square root of 98. Almost the square root of 100, which is 10 to the circle. And finally, we looked at the point 5, 9. This orange point out here. Zooming in we see that, yes, it's just outside of our circle.

## Shifting a Circle Right Four Points

Let's get back to our original circle with a radius of 50 units. Here's the circle, and then this is the center of the circle, 0,0. It's at the origin. Now, not every single circle we encounter will be centered at the origin. Let's try and move this one. Say we wanted to move the center 20 units to the right, to this point. This means if we move the center 20 units to the right. Then every other point on the graph must also move 20 units to the right. What would be the coordinates of these four points if we move them 20 units to the right?

## Shifting a Circle Right Four Points

Well, if we want to send each of these four points 20 units to the right, we just set 20 to all the x coordinates. So this first point at 50, 0 will now be at 70, 0. This point at 0, -50 will now be at 20, -50. And this point, we'll move to negative 30, comma 0. And finally the point 0, 50, we'll move 20 units to the right, so we add 20 to 0, giving us the point 20, comma 50.

## Shifting a Circle Right

Our new circle, is centered at the point 20,0 here and it has this equation. That's amazing thought if you got it correct. We know that for this blue circle, the radius still needs to be 50 units. This is why we still have 2500 on the right-hand side for our constant. The square root of this number is 50, our radius. Notice that for this orange circle and for this blue circle, the y values never changed. We're at the same height for any point along our graph. In other words, we only shifted the graph to the right. So, the only thing we really changed or manipulated were the x values. We moved all of our points 20 units to the right. So, the x values all increase by 20. To compensate for this adjustment, we need to subtract 20 from all our x values to maintain the same radius. So, for circles if we subtract something from x we really shift it to the right that many units. So, subtracting 20 moves this circle 20 unit to the right. So, if we add 20, we'll actually move our circle 20 units to the left. I really recommend playing around with this equation to see how you can move a circle around on a graph. Maybe see if you can make it go up or down.

## Shifting a Circle

Based on what you've seen so far, let's see if you can shift a circle. So here this orange circle was our original circle we saw with a radius of 50 units. This blue circle was our same circle shifted 20 units to the right. And keep in mind even though we went 20 units to the right, we see a subtraction of 20 in the equation I feel like this goes against my intuition. Usually when I think about moving to the right, in the x direction, I think about adding or positive this in mind when shifting a circle. What's in the equation will be the opposite of the direction we move. So if we have negative 20 in an equation, we really move our circle 209 units to the right. So, knowing this, what do you think would be the equation of a circle with center at negative 10, 2 and a radius of 50? Enter that equation here.

## Shifting a Circle

Here's the original circle we had with a radius of 50 and its center was at the origin 0,0. We want to have a new circle at center negative 10 comma 20. So we really need to move our circle 10 units left and 20 units up. This means in the equation we'll see a plus 10 next to the x. And a minus 20 next to the y. The positive 10 will shift our circle 10 units to the left, and we see that here. And the minus 20 will shift our circle 20 units up vertically. Notice too that the signs here are opposite. If we have a circle with a center at negative 10, then in the equation we have positive 10 next to the x. And if the center is at positive 20 for the y coordinate, then we have negative 20 in the equation. If you feel like you're stuck, try creating some other circles with other numbers and see how the circles move. Are you moving up, down, left, or right?

## Center of a Circle

What do you think would be the center of a circle with this equation? Enter that point here.

## Center of a Circle

This center would be at the point 4 comma 3. This is a graph of the circle, and we can see that the graph has been shifted 4 units right and 3 units up. This black circle is the equation x squared plus y squared equals 25. It's a circle with a radius of 5 units. And notice that it's center is at 0 comma 0. We'll we're really just shifting this equation. We're going to shift it right 4 units since we have minus 4 and up 3 units since we have minus 3. So, this is how our center shifts from 0 comma 0 to 4 comma 3.

## Identifying Parts of A Circle

So, this is the general equation for any circle. The number on the end is "r" squared, and so the square root of this number is the radius. And then if there's a number next to the "x", and a number next to the "k", these are the numbers that form our center. We take their opposite sign from the one that's in parenthesis. So, for example, if this is negative and this is negative, then both "h" and "k" will be positive. Whereas if there were positive signs inside of our equations, that means that these two values would be negative. Let's check out a few examples to see this in action. Here's a circle centered at the origin with a radius of 5. Notice that this equation is really the same as this one. If we subtracted or added zero to the "x" and "y" value, we'd still be at a center of (0,0). This is our "h" and this is our "k". I'm gonna switch between these graphs. Right now I only have this one being shown as graphed. When I turn this one off, and turn this one on, I notice that I get the same exact graph as before. And that should make sense because if we subtract zero from anything, it doesn't change the value. Here's another circle and this one has a center at (3,3). We took our old circle and shifted it three units to the right and three units up, vertically. We can even see that if we examine two of the points on the circle. This one, at (0,5) for this circle, has now been moved to (3,8), for this circle. We moved it three units right for the "x", from 0 to 3, and three units vertically from 5 to 8. And it turns out that this is true for all the points on this circle. We can examine another point and see that it was moved three units right and then three units up. And here's a different circle. This one's been shifted four units to the left, and then one unit up, vertically. And notice, too, that the center of this new circle is at (-4,1). This is the "h" value and it's opposite in sign of the equation, and this is the "k" value which is opposite of this number in the equation. Here's one final graph. For this one, it's been shifted two units left and two units down. And if we examine its center, it's at (-2,-2). Notice that these two signs are opposite than the ones in the equation. Here's our "h" value, and its opposite which is -2, and here's our "k" value, which is opposite, so -2. Now that you've seen the general equation for any circle, I want you to try and figure out what are the centers and radiuses for each of these circles? Work carefully, and don't forget to pay attention to the minus and addition signs. Good luck!

## Identifying Parts of A Circle

Here are the centers and here are the radii. Great work, if you figured each of them out. Let's start by focusing on the center. Remember the center is at h, k and we take the sign that's opposite and this parenthesis. So, we know the x coordinate here is positive 1 and the y coordinate of the center is positive 2. This is the h value. This is the k value. For this second one we'll have negative three for the x coordinate of our center. This is our h value. And we'll have positive 3 for the y coordinate of our center. This is the k value. And notice that since these signs match, we just take the positive 3 for k whereas here we have to have negative 3 for the value of h. This third one might have been tricky. We'll have positive 7 for the x coordinate of our center and we'll have 0 for the y coordinate of our center. There's nothing or subtracted in parentheses for this y, so we know the k value is just 0. And finally for the last one we'll have negative 4, 2 for this center. We change the sign here and we change the sign here. Now let's focus on the radius. If we take the square root of the number on the end, we get the radius. So, the square root of 16 equals 4, the square root of 1 equals 1, and the square root of 49 equals 7. For this last one, we'll take the square root of 8, which really equals 2 root 2. Now, you could have just written square root of 8 and that would have been correct as well. These two are the same number.

## Completing the Square for Circle Equations

Sometimes, though, the equation of a circle won't appear in this form. It could appear like this. When we see an equation with an x squared term, a y squared term, and other terms that have x and y at a constant, we want to re-arrange this so we can put it in this form. Notice that we have a perfect square of an x term, and a perfect square of a y term. If this is the case, we want to re-arrange this equation so we can group the x terms together and we can group the y terms together. We'll also want to move this constant to the right hand side of our equation. We do this so that we can complete the square for the x terms and for the Y terms. So what do we need to add to the left hand side of our equation to complete the square for these X terms. Keep in mind if we add an amount to this side we need to add the same amount to the right hand side of our equation so we keep the equation balanced. And, what do you think we would need to the y terms in order for this to become a perfect square. Again, whatever we add to the left, we need to add to the right.

## Completing the Square for Circle Equations

We'll add one to both sides to complete the square for the x terms. And we'll add 4 to both sides to complete the square for the y terms. Nice thinking if you got those numbers. Remember to complete the square, we need to add b over 2 squared. The b value is the coefficient in front of the single x term. Notice that we just have x to the first power. So that means b must equal 2 here. So we'll have 2 divided by 2 which equals 1 and 1 squared is positive 1. So we add this to the left hand side and the right hand side of our equation. For the y terms, our b value is 4. So we'll have 4 divided by 2 which equals 2 and we'll square it to get 4. This is what we add to the left and to the right.

## Rewriting the Equation in General Form

So now using this form, I want you to write this equation, so we have an x term squared. A y term squared. And then radius squared. What's our new equation?

## Rewriting the Equation in General Form

We can rewrite this trinomial as a perfect square, x plus 1 squared. And we can rewrite this trinomial as y plus 2 squared. And finally, we add the constant terms on the right hand side of our equation to get positive 9. This is our equation for our circle.

And now that we've turned this equation into one that's more easily recognizable, what's the center and what's the radius?

The center is at negative 1 comma negative 2 and the radius is 3, the square root of 9. Great work if you got these two right. Now, if you made a mistake, that's okay. Just keep in mind that the center is always the opposite sign of what's in the parenthesis here and here. We have positive 1 here and positive 2 here. So the h must be negative 1 and the k must be negative 2.

## Circle Practice 1

Alright, we have covered a lot of ground with circles. Let's see if you can answer your first practice question. What do you think would be the equation of a circle with a radius of 4 and is centered at the origin?

## Circle Practice 1

We'll want to have the equation x squared plus y squared equals 16. Great thinking if you got this equation. Remember that circles that are centered at the origin are of this form, x squared plus y squared equals r squared. We know that the radius is 4, so we simply replace r with the value of 4. So we'll have x squared plus y squared equals 4 squared, which is really 16.

## Circle Practice 2

How about this one? Which of these graphs, would be the circle for this equation? Is it this circle, this one, this one, or this one?

## Circle Practice 2

It would be this circle. We know this, since the center is at negative 2 comma negative 3. This is our h value, it's opposite in sign of this and this is our k value, opposite sign of the number next to the y in parenthesis. Great thinking, if you picked this circle.

## Circle Practice 3

Which of these circles would be the graph for this equation? Would it be this blue circle, this green circle, this orange circle, this black circle, this purple circle or this red circle? Choose the best one.

## Circle Practice 3

The graph would be this purple circle. We know this since the radius would be smallest radius, that would be 4. Our center will be at 2, negative 1. We know this since this sign is negative 2, so h is positive 2 and this sign is positive 1, so k is negative 1. The large circles had a radius of 16 and that would have been way too big. The next circles a size down had a radius of 8, still too large. So, we were really choosing between the smallest 2 circles. And it's this one because its center is at 2, negative 1. And if we zoom in, we can easily see that. The center is at 2, negative 1 here. Now, I know you didn't have this close up view for the circle, but the idea was to get an estimate of which one you think it is. Based on the direction that the center moved and based on the size of the radius. Sometimes it's just as good to have a qualitative understanding of about what your graph should look like, rather than an exact picture. That's what we were going for with this question.

## Circle Practice 4

Alright, let's see if you can tackle this one. What do you think would be the center and the radios for his equation? Keep in mind that this equation really represents a circle.

## Circle Practice 4

This circle will have a radius of 2 and a center at the point negative 3, 1. Great work if you found those two. Now if you struggled with this, I recommend going back to looking at completing the square. You'll need this as a helpful tool for the many of the types of problems, you'll encounter later. We want to start by subtracting 6 from both sides of the equation. So we'll have negative our y squared term and negative 2y, also on the left. I'm moving my constant term to the right. And I'm regrouping the x terms together, and the y terms together. We'll add 9 to our x terms, and we'll add that to both sides of our equation. We add 9 since b over 2 squared equals 9. B is 6. We divide it by 2 which equals 3 and we square it to get 9. To complete the square for the y terms. We'll add 1 to both sides of the equation. The b here is negative 2. So negative 2 divided by 2 equals negative 1. And then we'll square it, which equals positive 1. So these terms terms can be written as the perfect square of x plus 3 squared. And these three terms could be written as the perfect square of y minus 1 squared. And then on the right hand side, we simply add these constants together to get a total of 4. Now our equation is in a form we recognize. We have h, which is negative 3. And we have k, which is positive 1. And then our radius is simply the square root of this number, it's 2.

## Circle Practice 5

Try this one out. What do you think would be the circles' radius and the circles' center?