 I want to welcome you back to Episode 2 of Math 1050 College Algebra. I'm Dennis Allison here in the Math Department at Utah Valley State College. And today we're going to be talking about Cartesian formulas on the Cartesian plane and the equations of circles. But you know before we get into this let me remind you that there were 18 problems in the review pages on the website. You may have had a chance to look at that I hope by now. And remember if you look on the web page for Episode 2 you'll find the answers for all those problems and you'll actually see them worked out. So that's something you'll want to check out if you haven't done that yet. Also from Episode 1 you may recall that I mentioned one of the important differences between college algebra and intermediate algebra is college algebra focuses much more on graphs. Not completely, but there's a lot more graphing in this course than you saw in the previous course. And in the previous course it was more algebraic computation factoring real solving equations, things like that. But now we turn to something new. Now this course actually includes within it a lot of what you would call analytic geometry. Back in the 17th century the XY coordinate system was first introduced and that gave an opportunity to combine geometry and algebra. Prior to the 17th century geometry and algebra were considered two separate branches of mathematics that had a little or nothing in common. So for example if you would write an equation like this and showed it to someone prior to the 17th century, if you asked them what this is they wouldn't say that's the equation of a straight line because they wouldn't know anything about graphing. What they would say is this is an algebraic equation with two unknowns and it has lots of solutions. For example one solution is x equals 2 and y equals 2. Another solution is x equals 0 and y equals 4. Or another solution is x is equal to 5 and y is equal to negative 1. And that was as far as people got with that. Now if you mention lines people would think of lines as they were related to geometry. For example in a high school geometry course you study the relationships between lines and circles that lines together can make triangles and polygons and so forth. But lines were considered a strictly geometrical idea. Well along comes this guy named Rene Descartes. As a matter of fact we have a picture of him in the book we're using right now. He was born in 1596 and he died in 1650. He was French and back in those days his name was probably spelled D-E-S and then a separate word C-A-R-T-E-S. It was like since then it's been combined into one. Now the idea that he had was that when he studied mathematics he wanted to reduce things to the most simplistic form that he could. And he approached philosophy in the same way. You may have come across the same fellow in a philosophy course. In fact I believe someone told me before class they're taking an intro to philosophy was that in here? Yeah Matt, what have you heard about this fellow in your philosophy course? Well that he wanted to test truth by doubting everything and starting with the most basic concept that cannot be doubted which he had said Kōhito ergo sum, I think therefore I am, which was the most basic truth that he knew he existed because he thought. And he approached philosophy the same way he did with math by taking the most basic concept and building upon it. Right, exactly. I think in the discipline of philosophy I think René Descartes considered the first modern philosopher. In mathematics he's considered the first modern mathematician. And it's really for sort of the same reason that he tried to distill things down to these basics. In philosophy he began with what's the most basic thing I can know for example I exist. And he said well how do I even know that I exist? And his conclusion was that I think, therefore I am that is someone's asking this question about my existence. That must be, that must exist so he proceeds from there to build his philosophy. And many later philosophers begin in the same way. Can we really trust our senses? Can we trust what we see in here? Or can we be fooled? How many of you have seen the movie The Matrix? Okay so yeah, every one of you have. Well you know in that movie let's see who's the fellow, is it River Phoenix? The stars in that? I don't know. I've forgotten who that is. And Fishburne, yes. But you know in that movie the idea is that reality is not what it appears. And that he's in this so called matrix and he's being perceived his entire life is not what it seems to be. Well that's sort of right out of Cartesian philosophy that whole idea. Okay well here's the notion that René Descartes came up with interest to us. If you take a number line, a horizontal number line, call that the x axis, call another number line the vertical y axis, you can plot ordered pairs on here by going over like let's say if I go over two and up three, if I go over two and up three then I locate a point, then I locate a point right up there and we call that the point two three. I think you're already familiar with this idea. Now this is referred to as the rectangular coordinate system because well I can show you here because when you move to that point you move horizontally from zero zero the origin and you go up or you could go up three and then you could go over but basically you're moving around a rectangle so these are called rectangular coordinates. Also if I call this the x axis and if I call this the y axis then this is called the x y coordinate system but the last name for that coordinate system comes from his name. You see this is René Descartes I'm going to spell it separated like this and this is sometimes referred to as the Cartesian plane the Cartesian plane so it's named after him there. The way he describes how he came up with this idea is he said that he was laying in bed one morning he was accustomed to staying in bed late and sort of having a leisurely morning. He was laying in bed one morning he was looking up the ceiling and there were these square tiles in the ceiling and there was a fly that was landing on one place or another and it occurred to him for some reason I don't know that it would have occurred to me but it occurred to him how could I tell somebody where the fly is and he decided he could count so many spaces over and so many spaces up and he could tell someone an ordered pair to tell them where the fly is located and that was sort of the beginning of the idea that you could take ordered pairs of numbers and you could plot them and you might get who knows what. So when he took a linear equation like x plus y equals four remember we just looked at that a moment ago when he plotted the ordered pairs to his surprise the ordered pairs lined up they made a straight line I mean the average person might think that it would be sort of salt and peppered to the place that there would be no pattern to where the ordered pairs were but they actually made a geometrical shape and with other equations you could make the points become circles you can make the points become parabolas you can make them come to any other geometrical shape so that was sort of the beginning of this blend of geometry and algebra and this is what we call analytic geometry so I guess we could say René Descartes is the father of the analytic geometry and when we talk about graphing in this algebra course we're basically doing analytic geometry okay well let's look at some formulas that are related to the Cartesian plane and these formulas are referred to as Cartesian formulas. Now the first formula deals with this situation we have it on a graphic over here two points on the plane and to make it general I'll call this first point x1, y1 and I'll call the other point x2, y2 so x1 is the first coordinate, y1 is the second coordinate, etc and the question is how far apart are those two points that is what's the distance between them now you see the notion of distance is really a geometrical idea because to measure distance you'd think you'd need a yardstick or a ruler or a tape measure something like that and those are tools that come from geometry but René Descartes found a way of measuring that distance using algebra to answer a geometrical question. Here's how he did it if I pick the number down here on the x-axis directly below this ordered pair I think that would be x1 because I go over x1 before I go up and what number would this be on the y-axis what would we call that? That'd be y1, of course that'd be y1 and directly below the point x2, y2 this would be x2 and directly across from it on the y-axis would be y2 now here's the approach that he took he said if he drew a horizontal line across here and if he drew a vertical line across there and made a right triangle he could figure out the length of the hypotenuse if he knew the length of the base and the length of the height by taking the base squared plus the height squared equals the hypotenuse squared what theorem is that in geometry? exactly and you know that's one of the few theorems that actually relates algebra in geometry prior to this period because you have a right triangle and you have an algebraic equation a squared plus b squared equals c squared well so how long is the base? Well the length of the base would be the distance between x1 and x2 and generally the way we find the distance between two numbers is we subtract them x2 minus x1 but you know in a particular case x2 is bigger than x1 but in a different situation x2 could be smaller than x1 so I'm going to put absolute values on there to make sure that I get a non-negative answer for that distance and in the same manner what would be the distance from y1 to y2? y2 minus y1 y2 minus y1 and once again I'm going to put absolute values on there to make sure that that distance is never negative I take the base squared plus the height squared I get the distance squared let's see so right here this is the absolute value of x2 minus x1 over here this is the absolute value of y2 minus y1 and so let's work this out now on the green screen and see how that looks I'm going to take the absolute value of x2 minus x1 squared plus the absolute value of y2 minus y1 squared and that equals d squared you see this is actually an algebraic an algebraic equation that's representing that geometrical problem if I solve for d then what I would do would be to take the square root of both sides so I'll take the square root of this sum but when I write this down I'm going to drop off the absolute values and just say it's the quantity x2 minus x1 squared plus the quantity y2 minus y1 squared now what makes that legal by the way for me to drop the absolute values all of a sudden you're squaring them exactly we're squaring them you see we didn't know if this difference was positive or negative but in either case the square will be the same so there's no need to put the absolute values back in the problem at this point and the formula that I have right here is referred to as the distance formula I'll just put a box around that and that'll compute the distance between two points now here's the power of the formula we have a little space here so let's work a problem right here I'm going to pick two points maybe maybe you could pick points for me I want to point p and I want to point q somewhere on the plane let's see Matt tell us an ordered just any ordered pair three two three two okay and Susan what's another ordered pair one four one four okay now without even drawing an illustration I want to find out how far apart these two points are now you would think you'd need to actually see the points before you could say how far apart they are but since we're not actually going to measure them with a tool we're going to measure them algebraically I can find the distance between those two points using this formula now I need to decide which points going to be x1 y1 which one's going to be x2 y2 and you can actually do this either way let's say we call this point number one and we'll call this point number two then the distance will be the square root of let's see x2 minus x1 that's one minus three squared and you'll notice sure enough that difference is is negative this time but when I square it that won't be a problem and then the other quantity is going to be four minus two squared four minus two squared so the distance between these points is the square root of negative two squared plus two squared so the distance then is the square root of four plus four or the square root of eight now you remember last time we talked about simplifying radicals although technically this answer is correct if I simplify it how would I write that two root two two times the square root of two very good two times the square root of two now if you take out your calculator you could take the square root of eight or you could take the square root of two and then double it and you would see the same decimal appear but you know neither one of those decimal answers is actually accurate those are both approximations of the true distance this is the true distance right here or I guess you could say this is the true distance if I hadn't reduced it but if I took out a calculator and computed this I would get a decimal answer that would terminate after six, eight, ten decimal places depending on the calculator you use and the reason for that is when you have a square root and a number under the square root that's not a perfect square so you can't completely eliminate the square root these are called irrational numbers and the important characteristic one of the important characteristics of irrational numbers is in decimal form they never terminate and they never repeat a fraction or never repeat a decimal so I would leave the answer like this and not even use a calculator to approximate that okay let's take another another problem that pertains to these formulas what if I plotted two points or three points on the coordinate plane and let's see I picked out three ahead of time for us to plot let's take the point three two, three two is right here there's the point four negative one right here and the point negative two negative three right here now you notice those three points if I connect them make a triangle and it looks like it's close to being a right triangle but we don't know for sure I can use the distance formula to figure out if this is a right triangle let's see this is the point three two this is the point four negative one and this is the point negative two negative three what I could do would be to find the distance between these two points the distance between these two points and the distance between these two and I could check to see if the squares of the two shorter distances is equal to the square of the larger of the largest distance and if it is then by the by the converse of the Pythagorean theorem then this is a right triangle let's just try this on the green board and see what happens okay so we have a segment that goes from three two to the point four negative one and then we have another segment that goes from four negative one to the point negative two negative three and then from negative two negative three then we come back to the point three two so I want to check each one of those distances and see if they satisfy the Pythagorean theorem I'll call this distance d one d two and d three and distance d one is the square root of let's see I'll call this point number two four minus three squared and then negative one minus two squared and that is the square root of one squared plus negative three squared so it'll be one plus nine is the square root of ten and I can't reduce the square root of ten there aren't any squares in it and I don't want to use a calculator because I want to keep the exact answer to this problem now I'll go to distance number two this will be the square root of negative two minus four squared plus negative three plus one because I have to subtract a negative one negative three plus one squared and that's going to be the square root of let's see negative six squared is thirty six and negative two squared is four that will be the square root of forty and I can reduce that what would be the simplified version of the square root of forty two square root of ten two square roots of ten and now the last distance d three is the square root of three minus negative two let's just take a shortcut here three minus negative two is five squared and two minus negative three is five squared so what we have there is the square root of fifty twenty five and twenty five are there any squares that will divide fifty twenty five well sure so that's going to be five times the square root of two so here are the three distances and now we have to decide which one of those is the biggest well we have the square root of fifty is this one we have the square root of forty is this one and the square root of ten so it looks like square root of fifty or five square roots of two is the biggest so I'm going to be calling this distance c and I'll call these others a and b let's go back to the green board okay so from three two to four negative one we said that was the square root of ten from four negative one to negative three negative two negative three we said that was two square roots of ten and along the third side which we are wondering if that's the hypotenuse that's five square roots of two okay let's calculate or use the Pythagorean theorem to see if the equation satisfied I'll take the square root of ten squared plus the two square roots of two square roots of ten squared and the question is whether that will be the square root no five square roots of two squared let's see well how much is this quantity the square root of ten squared is what ten right that's what the square root means is if you square it you get ten and this one well if I square the two I get a four and if I square the square root of ten I get a ten so sure enough that's ten plus forty is fifty and over on the other side this is twenty-five times two twenty-five times two and that's fifty because that was the square root of fifty to begin with so sure enough those are equal and so this is a right triangle now in what branch of mathematics say back in high school in what branch of mathematics would you have seen right triangles more than in any other branch geometry in geometry ok so we found out that this is a right triangle in geometry but I didn't use any facts from geometry I used analytic geometry I used algebra to solve a geometry problem now in other cases you can use geometry to solve an algebra problem so that's the important aspect of the coordinate plane and analytic geometry is that you allow these two branches to really become one and they can play off each other ok let's go to a different problem that involves some new Cartesian formulas and this problem is to find the midpoint ok this time I have the same two points I had before x1, y1 and x2, y2 but the question this time is to find the midpoint now you know in a high school geometry course you do find the midpoint using a compass and straight edge if you remember what you do is you open up your compass you put the needle here and you make a big arc like that and then you flip the compass over you put the needle here you make an arc coming back this way then you draw a line through there and you find the midpoint it's a technique from a geometry course well I want to find the midpoint using algebra rather than geometry so what I'm going to do is I'm going to come down to the x axis x1 like we did before and this is x2 and the midpoint should be halfway between these two so the way you find oh that would be x bar so the way you find a number halfway between two numbers is you add them together and divide by two you average them for example if you were a student in one of my classes and you came in and you said Dennis on the first exam I made an 80 and on the second exam I made a 90 could you tell me what my average is well the average is the number halfway in between if there are two scores so it'd be 85 and so the average here will be the midpoint between x1 and x2 and the way I'll calculate it is I'll average x1 and x2 let's go to the next graphic and I'll show you what I mean you see there's a formula for x bar shown on the left and it's x1 plus x2 divided by 2 that's the middle number between x1 and x2 we'll do the same thing for the y's and y bar is y1 plus y2 over 2 so those are the two midpoint formulas let's just take an example of how we would work that suppose we were given two points let's say we're given a point P whose coordinates are 2, 3 and I'm given a point Q whose coordinates are 8, negative 5, 8, negative 5 now I could plot those two points on a Cartesian plane and I could connect them with a straight line segment and I could use a compass and straight edge to locate the midpoint but instead without even plotting them we'll be able to name the midpoint x bar, y bar that's the question what is the midpoint so to calculate x bar what I do is I average the original x's that's going to be 2 plus 8 all over 2, 2 plus 8 over 2 and that's going to be 5 and y bar is going to be 3 plus, let's see, plus 3 plus negative 5 all over 2 and that's going to be negative 2 over 2 is negative 1 so the midpoint has coordinates 5, negative 1 5, negative 1 right there ok so that's what we're going to use the midpoint formulas again in a moment but at least that gives you an example to see how I use it to calculate a point in the middle ok now I want to look at the equation of a curve on the coordinate plane a curve that you're familiar with but it's not a function of the equation of a circle let's suppose that, suppose that on the xy plane I choose a point to be the center of my circle and let's say this is the point h k and I use that for the center and around it I draw a circle I'll just kind of roughly sketch a circle right here it doesn't necessarily go right through the origin but it looks like I've drawn the origin and it looks like the radius here is r now if I have a set of points that form a certain curve I might wonder if there's an equation that all those points satisfy and there is in this case if I pick a random point on the curve let's say I pick a point up here for lack of a better name I'll call it xy I could have picked it here or here or here anywhere else but if it's on the circle there's one thing for sure I know about it and that is the distance from xy to the center is r just as it was over here now that's going to allow me to find the equation of this circle let's go to the green screen and we'll see how that's done I'll write that out the distance from xy to hk is r period now this is a sentence in English and I'm going to change it to a sentence in mathematics for example the word is translates as equals equals r and the distance from xy to hk well now we have a distance formula for that that would be x minus h squared plus y minus k squared so the distance is equal to r now you see what I have now is an algebraic sentence as opposed to an English sentence and when you hear people say that mathematics is a language this is exactly what they mean I didn't translate that sentence into Spanish or German or French I translated it into mathematics so maybe in some sense you should get a foreign language credit for this course because it is another language this is the equation of the circle except I want to reduce it and what would you suggest I do to make that look a little simpler on both sides okay so I have x minus h squared plus y minus k squared equals I have to square on the right too r squared and this is the equation of the circle but it's simpler than the one I had up above as a matter of fact this is referred to as the center radius equation of the circle and I think we have a graphic that has this on it let's go to that graphic yeah the center radius equation is x minus h squared plus y minus k squared is r squared and we've just arrived at formula let's work an example right below this on the screen suppose we had a circle whose center was at 3 negative 1 and let's say its radius was 5 we know the center we know the radius so that circle should be as they say well defined there's only one such circle with this center and this radius so what is its equation well its equation would be x minus 3 because h is 3 here plus y minus negative 1 what am I going to put here y plus 1 right y plus 1 squared equals 25 that's the equation of the circle now if I were to multiply that out I could make the answer look different but it's still equivalent to it I just don't see the center and I don't see the radius squared over there if I multiply this out this would be x squared minus 6x plus 9 plus y squared plus 2y plus 1 equals 25 is that large enough for you to read can you read that ok ok and if I rearrange terms I have x squared plus y squared minus 6x plus 2y let's see now I have a plus 10 and if I bring the 25 over that would be a minus 15 equals 0 if I write the equation in this form this is called the general or the standard equation of a line you notice I no longer see the center like I could see the center before I no longer see the radius so this has merely been expanded and in some sense simplified depends on your point of view what simplify really means so now let me take another example that involves the center radius equation and the standard equation I think we have another graphic coming up that has the general equation on it let's go to the second yeah there it is so as a general rule would say that the general equation can be written in the form x squared plus y squared plus a x plus b y plus c equals 0 a b and c are just some real number coefficients what if I gave you this equation of a circle x plus 4 squared plus y minus 6 squared equals 4 who can tell me the center of that circle negative 4,6 very good negative 4,6 it's not 4, negative 6 because h and k were subtracted so negative 6 was subtracted here 6 was subtracted oh I wrote it backwards didn't I yeah negative 4 positive 6 yeah I got it exactly backwards negative 4 positive 6 is the center and what is the radius 2 2 yes because this is the square of the radius exactly what if we take what if we take this circle x squared plus y squared plus 2x minus 3 equals 0 now what's the center and the radius of this equation well of course the problem is it's been expanded and changed so we can no longer see the center and the radius so I think if we could get it back in the form of the previous problem we could read off the answer to do that I'm going to put all the x's together and I'm going to put all the y's together and the 3 I'll move out of the way I'll bring over here to the other side now I want to complete the square what number should I add on here to complete that square to make this a trinomial square do you remember how that goes you get the middle and you divide it by a half and then time and square it yeah you take the middle number and you multiply it by a half or divide by 2 and you square that now if I divide 2 by 2 I get 1 and when I square it I get a 1 but if I add a 1 there I need to add a 1 over here I'm kind of running out of room so I'll just put a plus 1 right below it there same thing here if I take half of this coefficient or if I divide it by 2 half of negative 10 is negative 5 if I square it I get 25 and then I'll need to add 25 over here normally I'd write that out horizontally but I'm kind of press for space so this is now a perfect square what is this the square of x plus 1 x plus 1 squared very good and what's the other one squared let's see Jeff what's this other one squared y minus 5 squared y minus 5 squared yes and that's going to equal 29 now can you tell me the center and the radius for this circle Matt what would you say I'd say that the center is negative 1 5 negative 1 5 and the radius is 5 let's say square root of 29 yes so it turns out in this case the radius is not a nice integer it's the square root of 29 without a calculator could anyone tell us like between which two consecutive integers this flies between 5 and 6 between 5 and 6 yeah the square root of 25 is 5 the square root of 36 is 6 would you guess this is closer to 5 or closer to 6 closer to 5 yeah tell us why you think oh because it'd be close the square root of 25 is 5 and the square root of 36 is 6 it's closer to 25 than it is to 36 29 is relatively closer to 25 than it is to 36 so if I were going to guess I would say this square root is probably about 5.4 somewhere in that ballpark you could use a calculator to check it but we'll leave our answer like this because this is the exact answer now if I plotted a circle located the center at negative 1 5 and if I made a radius a little bit over 5 maybe 5.4 5.5 it would be kind of hard to estimate that if you draw it especially freehand but if you draw that circle you are drawing a graph of this equation right here let's take another circle problem this time I'm going to draw a circle and I want you to tell me what its equation is so let's use let's use a let's use a graphic that I've already made up here's here's an underlying sheet for it okay I'm thinking of a circle and the center is right here and the circle looks like this okay I want to find out what's the equation of that circle now you can think of the equation of a circle is being like the name of a person my name is Dennis my name is Mr. Allison you have names people call you by different names at home at school and so forth but they all refer to you and this circle has various names but they're both equations there's the center radius equation and there's the general equation I think probably the center radius of the equation is the one that would come up with what's the center of this circle 3-3 looks like it's 3-3 yeah 3-3 and what's the radius of this circle 3 looks like the radius is 3 because it looks like it comes up tangent right there and also tangent right here those are both 3 units so I bet the radius is 3 all the way around so it's a circle so its equation would be what Jeff x-3 squared plus y-3 squared equals 9 yeah x-3 squared plus y-3 squared equals 3 squared r or 9 so that's the name of this circle if it wanted to take the time to multiply it out in combined terms and put everything on one side that would be the general equation but I don't think we need to do that right now okay let's move on to another problem what if what if I were given a function and the function rule is 2 minus x now if I were to graph that what would it look like Jeff what were you going to say? It could be a line going through the y-axis where y is equal to 2 and with the negative slope yeah it would have y intercept 0,2 and it would have slope negative 1 it may look a little foreign because I have these written out of order I could turn it around and say negative x plus 2 so when we see this equation we know it's a straight line graph and we know exactly how to graph it what if I change the equation to this suppose I say I'll call this one g of x and this is the square root of 9 minus x squared well that's obviously not a linear function but it looks like it's a function we have a function rule what would you say is the domain of this function well let's see now the number under the radical can't be negative so if I come over here I know that 9 minus x squared has to be 0 or larger so I'm going to solve this inequality this tells me that 9 is greater than or equal to x squared and if x squared is smaller than 9 that means x is pinned between two numbers what is it pinned between negative 3 and negative 3 right very good you see x's can be negative but when you square them we just have to make sure we get no more than 9 so it could go down to negative 3 or 3 so I know that the domain of this function is the closed interval negative 3 2 3 now I want to draw a graph of this function but I've never graphed it before here's how I would do it I would replace the g of x with a y those are actually sort of interchangeable the y coordinate and the g of x value and because I've not seen this before I want to see if I can make it convert it transform it into a form that I would recognize so I'm going to square both sides y squared is 9 minus x squared and does anybody recognize that I don't think we've seen that before what are you thinking Jeff? it looks like the equation for a circle if you move the x squared over to the other side oh yeah if I move the x squared to the other side that is if I add an x squared to both sides I get x squared plus y squared equals 9 now that's beginning to look like the center radius equation of a circle I might even write that as x minus 0 squared plus y minus 0 squared equals 9 this is a circle whose center is at the origin and whose radius is 3 but you know a circle is not a function because a circle doesn't pass the vertical line test there are vertical lines that intersect the circle more than once but you know if I go back to this form or this form up here you notice why can't be negative because it's the positive square root that is the non-negative square root of 9 minus x squared so what this tells me is it's the upper half of the circle I think I can just sketch this right over here on the side if I go out to 3 and if I go back to negative 3 that represents the domain and if I draw a semi-circle of radius 3 so that makes this 3 up here that is the graph of the function g I'll just put a g beside it what would you guess would be the equation or the function rule for the lower half of the circle if I wanted that instead yeah Steven? Put a negative in front of the radical? yeah if I had taken the negative square root I would have gotten the lower half of the circle when I take the positive square root I get the upper half of the circle so even though a circle isn't a function a semi-circle can be a function and I find out that this isn't just a semi-circle very good okay this time I want to demonstrate an example of an inequality what if I draw a circle let's see I think I have another graph I can use here what if what if I draw this circle but I'm going to shade in a region let's say I pick this point as the center and I'm going to make this dotted right here that's supposed to be a circle and I'm going to shade in the interior now when you have a region sketched you can sometimes represent that using an inequality and what I will do is I will write the equation of the circle as if that were a solid circle with no interior shaded and then I'll decide whether I want the interior or the exterior by converting that into an inequality can anyone tell me the equation of the circle around this let's see what we need to know it's center what's the center 1, 2, 3, 3, 0 yeah 3, 0 and the radius is looks like the radius is 2 so the equation of the circle if I had drawn a circle would be x minus 3 squared plus y minus 0 squared I'll just put a y squared equals 4 but you know if you want the interior of a circle then what you do is you make this x minus 3 squared plus y squared is less than 4 it's now an inequality because if you remember when we first started talking about circles this was the distance from x, y to the center squared and this was the radius squared and this time if I pick a point like right here in the interior let's say I call that point x, y because it's on the interior of the circle this distance is less than 2 and so when I square both sides this is the square of the distance and this is the 2 squared the distance squared is less than the 2 squared what would I do if I wanted the points outside the circle out here greater than 4 yeah would make that greater than 4 and what would be the difference if I put an equal sign right there less than or equal to 4 what would be the difference now it would be a solid circle plus the interior because this is now the distance squared can equal 4 or be less than 4 so points on the circle and inside but since we didn't have that I'll leave that out okay what if I turn this around I'm going to give you an inequality and I'm going to ask you to sketch a graph of it so or we'll sketch a graph of it together so let me write an inequality up here such as x plus 1 squared I'm going to put a 4 in front of this one 4 times x plus 1 squared plus 4 times y minus 2 squared is less than is less than or equal to let's say let's say 1 what if I put a 1 there well of course these 4 sort of make it look different first thing we might do is just get rid of the 4's I'm going to divide out the 4's so dividing by 4 I get a 1 fourth over on the right hand side but this is beginning to look like an inequality that involves a circle the center of the circle is what negative 1 2 good and the radius a little different from the other examples the radius is what? 1 half you see this is the square of the radius so the square root of 1 fourth is 1 half right so when I go to graph it I think I'll just draw this one free hand up here when I go to graph it I'll locate the point negative 1 2 here's negative 1 2 right here and I'm going to go out the unit which means this is a pretty small circle I'm going to draw when I do this in the inequality should I make it solid or dotted? solid I think because it says less than or equal to so if I go out 1 half unit we'll say that's 1 half unit right there and I'm not quite done what do I still have to do? shade the inside have to shade the inside sure so I'll shade the inside and so the center actually satisfies the inequality and if that had been greater than or equal to 1 fourth I would have shaded the outside of the circle okay very good okay now class in summary I think we have one more circle problem that is on the screen now that sort of summarizes some of the information that we've just looked at we're given a circle it's centered at the point 1 negative 6 and the radius is 4 and I have 3 questions here to answer first of all find its center radius equation then find its general equation but then there's a part C why don't we come back to part C once we've established parts A and B so let's just see the center here is at 1 negative 6 and the radius is at 4 part A wants us to find the center radius equation Jeff what would the center radius equation be in this case? X minus 1 squared right plus Y plus 6 squared Y plus 6 squared equals 16 that's the answer for part A that's all it takes now for part B it said find the general equation you remember that's a little bit different form Susan how would I find the center the general equation just by multiplying it out okay what would be X minus 1 squared X squared minus 2X plus 1 okay and Y plus 6 squared I'll do that one is Y squared plus 12 Y plus 36 equals 16 you remember when you square a binomial you square the first term you square the last term but in the middle a lot of times students forget the middle term it's this product doubled it's 6 times Y doubled or 12 Y now if I regroup the terms I get X squared plus Y squared minus 2X plus 12 Y and let's say I need to accumulate some constants if I'm going to set it equal to 0 I've got a 1 a 36 and a 16 I have to subtract off so what constant would go here? 21 21 yeah 21 so this is the answer for part B this is the general equation or the standard equation of the circle now let's go back to the graphic there was one more question part C says is the point negative 2 negative 3 inside outside or on the circle well you know if I substitute that point into either of these equations if I get the two sides to be equal that means the point is actually on the circle but if I get an inequality it's going to be either inside or outside I think what I'll do is substitute that point into the center radius equation so if we come back here I'm going to erase this second answer so that I can use the first answer now the point we were given I'll call it point P is negative 2 negative 3 I'm going to plug in negative 2 for X so I get negative 2 minus 1 squared and I'll plug in negative 3 for Y that's going to be negative 3 plus 6 squared now the question is do I get 16 or more or less negative 3 squared or 9 this is 3 squared or 9 what I get is 18 now I claim that means the point is actually outside the circle because you see what I've done is to calculate the square of the distance from point P to the center this is the square because I didn't take the square root of that and so the distance from point P to the center is actually the square root of 18 not 4 more than 16 when I squared it so my conclusion is that point P is outside the circle now I mention this because if you draw a rather accurate graph of the circle that we started with center at 1 negative 6 radius 4 it's going to be rather difficult to tell if negative 2 negative 3 is on the circle or not if you draw it sort of free hand it's kind of hard to tell because it's very close to the circle but we've determined it's actually outside ok one more problem there's a graphic called David and the big guy ok you're probably familiar with this story in the bible David and Goliath meet and David has a slingshot and he throws a stone at Goliath so I have David located at the origin the circle going around the origin represents the path of the slingshot and at the point 5 1 the stone is released and it continues along the tangent line and it hits the big guy on the y axis and the question is how far away is David from the big guy now this actually doesn't require that we know the equation of a circle actually this is more closely related to equations of lines but there's a circle in the problem so I decided just to stick it in here at the end of this and I think it's a good application of mathematics to solving a quote real world problem so I'm going to draw that illustration right over here I have a circle I find the point 5 1 5 1 the stone is released and it's going to cross the y axis right there but we don't know what that point is it's probably further up than what I've shown it but I don't have a lot of room here well what's the relationship between the path of the stone and the circle well actually I think I just mentioned that a moment ago it's along a tangent line now what's the relationship between the tangent line and the radius to the center how are they related Stephen they're perpendicular they're perpendicular very good exactly so you know what if I could find the slope of this radius to the point 5 1 I could figure out the slope of the path of the stone what is the slope of that radius right there 1 5th it's 1 5th yeah because it looks like if we go over 5 and up 1 the rises 1 the run is 5 the slope is 1 5th now with that information I can figure out the slope of the path of the stone what would it be negative 5 negative 5 yeah technically negative 5 over 1 but we could reduce it to be negative 5 now let's see I think with that information I can figure out what is the actual equation of the path of the stone let's see what do I know I know the slope of the line and I know a point that it goes through so what equation from the last episode do I want to use here which linear equation the point slope equation and how did that go y minus y1 equals m times x minus x1 very good where x1 y1 is a point on the line well you know we have a point on the line the point is 5 1 so this would be y minus 1 equals m times x minus 5 but we also have m don't we so let's plug that in y minus 1 equals negative 5 times x minus 5 now you might say well Dennis that's well and good but how are we going to figure out where the big guy is up here just by knowing that anyone have an idea what we should do now Stephen convert it to the slope intercept form and how would that help because the constant that's being added the b is the y intercept that's the y intercept and that's exactly where the big guy is so let's solve for y y minus 1 equals negative 5x plus 25 so y is equal to negative 5x plus what 26 plus 26 well once again the slope is negative 5 we already knew that but what's important now is that b is 26 now that's what we were after so what that tells me is he's at 26 on the y axis and so how far is David from the big guy 26 feet he's 26 feet away so the answer is 26 feet exactly well you know what this problem does is it doesn't really demonstrate properties of circles so much except it does use some geometry but more importantly what it demonstrates is characteristics of straight lines and I use the property that the tangent to a circle is perpendicular to the radius that's actually a theorem in high school geometry which you may or may not remember Stephen remember because he told us that those would be perpendicular but we also use the fact that two lines that are perpendicular have slopes that are negative reciprocals of one another that was an idea from the previous episode and that's the point slope equation of a line and we use the slope intercept equation of a line let me ask you another question about this problem how long is the rope on the slingshot here's David he's whirling a stone around his head how far out is the stone as it goes around David well I think that would be the radius of the circle wouldn't it and we know a point on the circle so how can I get the length of the radius from five one to zero zero where David is using a formula from this episode distance formula very good the distance formula okay so the distance which is the radius would be five minus zero squared plus one minus zero squared and I think when you work that out you get the square root of twenty six so it's a little over five feet well class let's see what we've talked about today were several Cartesian formulas there's the Carti there's the distance formula there are the two midpoint formulas x bar y bar then we talked about two equations for circles there's the center radius equation and there's the general equation of a circle and then we looked at a number of problems related to that when we come back next time we'll talk about graphing on a graphing calculator we'll talk about plotting points to graph curves and we'll talk about some fundamental graphs I'll see you then