Hey everyone, and welcome back. Let's see if we can give this problem a try. So in this example, we're asked to write an equation for a circle with each of the following characteristics. We're going to start with situation A. Now, before I do anything, what I'm going to do is write the standard form for any circle, which is:
x-h + y-k = r2This will be how I write the equations once I find what the graph looks like. Now we're going to start with this first circle here. You can see our center is at 1, so that means horizontally we're at 0 and vertically we're at 1. So I can put h is 0 and k is 1, and that's going to be our center. Now we're also told that our radius is 3. So what that means is I can go 1, 2, 3 units to the right; 1, 2, 3 units to the left; I can go 3 units up, and I can go 1, 2, 3 units down. And this is going to give us our circle, which is the shape that we're looking for. So this is what the graph looks like, and what I also need to do is write the equation. So to find the equation, I can use the standard equation, which is going to be:
x2 + y-1 9And that is going to be the equation and graph for this circle. Now for our next circle, we're given the center at 1, -2, and this has a radius of 1. So what I can do is go to our graph, and horizontally we're going to be at 1, and so that's going to be right here, and then vertically we're going to be at -2, which is down here. So that's the center of our circle. I can see that our radius is 1, so I can go 1 unit up, down to the left, and to the right, and this is going to give us the circle we need right here. Now in order to find the equation for this circle, I can use this same standard equation. So we're going to have:
x-1 + y+2 1And this is the equation we're looking for for this circle. Now for this last situation that we have, we have a center at -6, 3 for situation C, and then we have our radius as the square root of 5. So first, I'm going to find our center on this graph, which is going to be horizontally at -6 and vertically at 3. So if I go to the left, I'd be at -6, and then I can go up to 3, which would be right about there, and this is the center. Now this one's going to be a little tricky, because our radius is the square root of 5, and the square root of 5 is approximately 2.2. So what we can do is look at where the center of our circle is, and we know that vertically the center is at 3, right. So what that means is I can go 2 units down which would put us down here at about 1, but I need to go a little bit past that which is going to be right here, and that would be one of the points. Now since we're vertically at 3, I need to go up 2.2 units which would put us here at 5, but again since I'm it's 2.2, I need to go a little bit past that to be right about there. Now what I can also do is go to the left and to the right. Now I'm gonna go to the left 2 units, so we'll go from -6 to -8, which is right about here, but I need to go a little past that since it's 2.2, and I would go from -6 to -4 to the right, but again go a little past to put a point there. And this is going to be where the points line up, and that is how you can draw this circle. Now the equation is actually pretty straightforward. So to draw the equation, all you need to do is use the standard equation, which would be:
x+6 + y-3 5And this is going to be what the equation looks like for our circle, and that is how you can find the graphs and equation for these circles based on the characteristics that we were given in this problem. So, I hope you found this video helpful. Thanks for watching.