[Music] here's lesson three of the second unit this lesson is on the equation of a circle let's start by looking at the definition of a circle a circle is the set of all points that are the same distance from a fixed point which we call the center of the circle the radius is the distance from the center of the circle that fixed point to any other point on the circle what I'm going to do next next is show you the equation of a circle that is not centered at the origin and the equation of a circle that is centered at the origin and then after I show you the equations I'll give you a detailed explanation as to why the equation is able to describe the relationship between any XY point and the radius of the circle so first of all let's start with any Circle so the circle that could be centered anywhere doesn't have to be at the origin in that case we call the center of the circle Point h k and the relationship between any XY point and the radius is x - h^ 2 + y - k^ squar is equal to the radius squared if the circle is centered at the origin that would make the H and K values zero which means the equation would simplify to just x^2 + y^2 = R2 now why do these equations describe the relationship between any XY point in the radius well maybe let's look at a more clear diagram of a circle here I have a circle drawn that's centered at the origin so I have the center labeled as the point 0 0 if you can imagine me drawing a radius from this Center 0 0 to any point on the circle so anywhere on the circumference of the circle it would connect to the circle at some point XY so I'll label this point on the circumference of the circle as Point XY and I've drawn the radius let me label that as R to represent the radius of the circle now if I were to describe the relationship between this XY point and the radius if I were to go from the origin to that XY point I could travel X units parallel to the x axis and then y units parallel to the Y axis so let me label those side lengths as X and Y and notice what I've created here is a right angle triangle and what do we know about the relationship between the sides in a right angle triangle we know that the sum of the squares of the shorter two sides so x^2 + y^2 is equal to the square of the longest side in this case the longest side is the radius so it's equal to the radius squar and this is the equation of a circle x^2 + y^ 2al R2 and that relationship between the XY point and the radius holds true for any XY point on the circumference of the circle if the circle is centered at the origin so hopefully now you can understand where this equation of a circle comes from let's now practice writing the equation of a circle example one says write the equation of a circle with Center 0 0 and a radius of in part A3 so it gives us the radius and when we write the equation of a circle remember we write it in the format of x^2 + y^2 equals the radius squar and in this case we know the radius is 3 so we would write it as 3^ s and whenever we're expressing our final answer for the equation of a circle we'll need to simplify the radius squared so we'll need to actually evaluate what 3^ s is so in this case the equation of this circle would be x^2 + y^2 = 9 and that'll be our final answer for the equation of that Circle let's try another one this time the radius of our circle is a half so when I write the equation of this circle it would be x^2 + y^2 equals a half squar and remember when the base of your power is a fraction what we need to do is apply the exponent to the numerator and the denominator so I need to square the one and square the two so 1 squar is 1 and 2^2 is 4 so the equation of this circle is x^2 + y^2 = a/4 let's move on to example two where this time it gives us the equation of the circle and it asks us for what the radius is example 2 says what is the radius of a circle defined by the equation x^2 + y^2 = 36 first thing you need to remember is the number that you see on the right side of the equation is what the radius squared is equal to so in this case the radius squared is 36 if we're interested in the radius we'll have to isolate r on the left R is being squared so to isolate R we have to do the inverse of squaring which is square rooting so R would be equal to the Square < t of 36 which is 6 so the radius of this circle is 6 units example three says a circle has a center at the origin and passes through the point 53 determine the equation of the circle so this time it gives us an XY point that is on the circumference of the circle so in order to determine the equation of this circle let me start by writing out the general formula for the equation of any Circle centered at the origin x^2 + y^2 = R2 in order to write the equation of this specific Circle I'm going to have to solve for what R 2 is equal to to do that I will sub my XY point in for X and Y that will give me 5^ 2 + 3^ 2 is equal to R 2 5^ s is 25 3 2 is 9 so I can see that R 2 is equal to 34 now I'm not actually interested in what R is equal to I don't care what the radius is the question doesn't ask for it just wants the equation of the circle and in the equation of the circle I just need to know what the R 2 value is and I have it it's 34 so I can write my final answer x^2 + y^2 = r 2 which is 34 so this is the equation of the circle centered at the origin with 53 on its circumference the next question is a very typical question for this section of the unit it wants to know if a point lies inside outside or on a specific Circle let's look at the general rule for determining if any given point Falls inside outside or On Any Given Circle if the given XY point is on the circle then x^2 + y^2 would be equal to the radius squar that's given in the equation of the circle if the XY point is outside the circle then x^2 + y^2 would be greater than the given R 2 value and if Point XY is inside the circle then x^2 + y^2 would be less than the R 2 value that's given by the circle let's try it out for the information given in this question it says is the point -59 inside outsider on the circle x^2 + y^2 = 100 what we have to do is sub in the -59 for X and Y into the equation of the circle that we have and see if x^2 + y^2 is less than equal to or greater than the 100 value that is the r 2 of this circle so we're checking what is the relationship between -5^ 2 and 9^ 2 in comparison to this 100 value that is the r 2 value of this circle is it greater than less than or equal to so I just put an equal sign with a question mark above and we're going to test out to see the relationship between the left and right side of this equation on the left I have 25 + 81 and how does that relate to 100 well 25 + 81 is 106 I know that is greater than 100 that tells me that the point -59 is actually outside of the circle that's given let's verify that in Desmos so if I graph the equation of the circle x^2 + y^2 = 100 we have it right here and if I were to plot the point -59 we figured out that its x^2 + y^2 value would be bigger than 100 it was 106 so that should tell us that it's outside the circle and if I plot it notice that that point -59 is just outside the circle so we got the correct answer we should summarize our answer by saying therefore the -59 is outside of the circle let's move on to question number five this is the first time we're finding the equation of a circle where the center is not at the origin so let me remind you what the equation of a circle not centered at the origin looks like it is x - h^ 2 + y - k^ 2 = r^ 2 H and K that you see in this equation is the center of the circle so in this case 34 are our H and our K values and it tells us the radius R is 8 so to write the equation of the circle I'll just plug in my h k and R and then I'll have the equation the answer would be x - 3 2ar + y - 4 2ar equal 8^2 and remember when writing the final answer for the equation of a circle we need to simplify the radius squ so I'll actually have to evaluate 82 my final answer will be x - 3^ 2 + y - 4^ 2 equal 64 and now let's go on to the last question of the lesson it says to determine the shortest distance from the point 107 to the edge of the circle x^2 + y^2 = 49 now you can see in this diagram I have the 10 7 labeled and I have the graph of the circle x^2 + y^2 = 49 now I've given you this so you can visualize what we need to do and then we'll algebraically calculate it and I put an important tip here it says that the shortest distance from a point to a circle is always going to be along a line that goes from the point along a line that goes through the center of the circle so the first thing we're going to do is calculate the from that 107 to the center of the circle and we can do that using your distance formula so I'll start by finding the distance from the point 0 0 which is at the center of the circle to the point 107 and using the distance formula I know it will be equal to the square root of the difference in the x coordinates squared so 10 - 0 squared plus the difference in the y-coordinates squared so 7 - 0^ SAR if I simplify underneath the square root I would have 10^ 2 + 7 SAR that's 100 + 49 which is 149 that has an approximate value of 12 something but we're going to keep the exact value for now so I know the distance of the entire line from the 107 all the way to the center of the circle I don't want that entire length I just want the portion of that line that goes from 107 to the circumference of the circle I could figure out that portion of the line if I subtracted the radius of the circle from the otk of 149 which is the length of the entire line so let's find the radius of the circle since the equation of the circle is x^2 + y^2 = 49 I know that the R 2 value is 49 and then to calculate the radius I would do the inverse of squaring which is Square rooting and I can solve for the radius to be 7 units long so I know the entire length of the line we see in the diagram isun 149 I know the radius is s if I find the difference in those two that will give me the portion of the line segment I'm looking for which is the shortest distance from 107 to the circle so the shortest distance is equal to the square otk of 149 minus 7 and we can get an approximate value for that with our calculators to two decimal places it is about 5.21 units that's our final answer and that brings us to the end of the lesson make sure you go to Jensen math.ca and try out the practice problems