Module 7: Conics Lesson 1 Notes Part 2: Circles Video 2 Transcript Hey everyone this is a video for module 7 lesson 1, circles, notes part 1 and this is the 2nd video. So the first example in this video is, what is the equation for the circle that has a center of (-­‐3, 1) and goes through the point (-­‐1,-­‐2). So the first thing I want to do is sketch this circle here. So I know that the center is (-­‐3, 1) which is here and I know that it passes through (-­‐1, -­‐2). So I know this line segment would represent the radius. So to get the equation of this line I know that it is “x” minus “h” squared plus “y” minus “k” squared equals “r” squared. And since I know that (-­‐3, 1) is the center, then I know that this could be “x” minus -­‐3 squared plus “y” minus 1 squared. So all I really need to know is my radius and that would complete my equation here. So I am going to clean this up just a bit. And now I have to solve for my radius squared. Now in your notes, you seen that we used the distance formula to find the length of this radius. So if I have a point (-­‐3, 1) and I have a point (-­‐1, -­‐2) I can use the distance formula with these 2 points to find the length of that. And that distance formula is really just embedded into this equation, I also showed you can get the Pythagorean theorem in your notes from the distance formula. So you know that it goes through (-­‐1, -­‐2) I can actually just plug those values in here. So negative 1 can go in for this “x” and then -­‐2 can go in for the “y,” since (x,y) is an ordered pair the circle goes through, I just plug it in and then solve. So notice this is really just x2 minus x1 squared plus y2 minus y1 squared, and then to get the radius I am going to have to square root that, which is the same thing as finding the distance. So this is going to reduce to 2 squared and this right here will be -­‐3 squared equals the radius squared. So that will be 4 plus 9 equals r squared. Which that would equal 13. So 13 equals “r” squared. So I had to find the equation of this circle, so if I am looking for the equation then I know that I am trying to fill this is. So I could go ahead and find the radius but since this equation has “r” squared in it, I am just going to go ahead and plug 13 in for “r” squared. So that is going to be, “x” plus 3 squared plus “y” minus 1 squared equals 13. And that’s the equation of that circle. Ok, the next question says change the equation of this circle, 𝑥 ! + 18𝑥 − 4 + 𝑦 ! −
6𝑦 + 10 = 19 into the standard from of a circle. So one of these are shown in your notes, and you had to complete the squared twice in this problem. So go ahead and look back in your notes if you need to. I am going to show you how to do it as well here, but there are detailed steps in the notes. So what you want to do first in this problem is get all the numbers to the right-­‐handed side of the problem. So I am going to move the 4 and the 10, I am actually going to put those together, so that would be a positive 6 over here, to move that over I would subtract, so if I subtract 6 from that side, that would give me 13. And I am going to put “x” squared plus 18x and then plus “y” squared minus 6y over here. All right I left these blank for a reason. On both sides I am going to add half of the “x” term squared and then half of the “y” term squared. So here I would add, half of 18 and square that. If I add this to the left side I would have to add it to the right side to keep it equivalent. So that would be 18 over 2 squared. And here I would add half of this, so half of -­‐6 squared, and then I would add that same value over here. Now, I am just going to simplify that. So on that side that would become “x” squared plus 18x, 18 over 2 that would be 9, 9 squared would be 81. Plus “y” squared minus 6y this is going to be -­‐3 squared is 9. So on this side I would have 13 plus 81 plus 9. Now I am not going to put the 81 and the 9 together, I am going to factor each trinomial here. And if you notice I have made a perfect squared trinomial there, where this is going to factor to (x+9)(x+9) by taking half of the “b” term essentially and squaring that, that allowed me to break it down into the same 2 binomials. Here it’s the same thing, so this would break down into (y-­‐3)(y-­‐3). 13 plus 81 plus 9, 81 plus 9 is 90 plus 13 would be 103. So if you notice now, I can write this as “x” plus 9 squared plus “y” minus 3 squared equals 103. I have changed this equation to the standard from of a circle. You can see now that my center is negative 9 positive 3 and I would have a radius of the squared root of 103.