Current time:0:00Total duration:6:16
0 energy points
Studying for a test? Prepare with these 12 lessons on Conic sections.
See 12 lessons

Common tangent of circle & hyperbola (1 of 5)

Video transcript
The circle x squared plus y squared minus 8x equals 0, and the hyperbola x squared over 9 minus y squared over 4 equal 1 intersect at the points A and B. Equation of a common tangent with positive slope to the circle as well so the hyperbola is-- so let's just visualize what they're asking first. And this is going to take us multiple videos, I have a feeling. But let's just visualize it just so we can get our head around the problem. So this circle, let me complete the square in terms of x. So this circle, as they wrote it, is x squared minus 8x. And then we have a plus y squared-- I left some space here so we can complete the square-- is equal to 0. And then let me add half of this 8 term squared to both sides. So half of negative 8 is negative 4, and negative 4 squared is 16. So add 16 to both sides. And that allowed me to turn the x-term into a perfect square. This is the same thing as x minus 4 squared. And then we have plus y squared. Plus y squared is equal to 16. So this is a circle. This right here is a circle with center at x equals 4. At x equals 4, y is equal to 0. And it has a radius of 4 as well. So let me graph this circle here. So let me draw the horizontal axis, my x-axis. Let me draw the y-axis. That is my y-axis over here. And let me draw its center, so 1, 2, 3, 4. 4 comma 0. That's it's center, and it has a radius of 4. So it's going to come out, and it's going to look something-- I could draw a better circle than that. It's going to look something like-- that's the top half, and then the bottom half is going to look something like that. So that's our circle. Now let's think about the hyperbola. So if we just look at it, the x squared term is positive, so it's going to be a hyperbola that opens to the right and the left. We do this a bunch in the conic sections videos if you want to review of that. And we could just figure out where it intersects the x-axis. So then when y is equal to 0, we have x squared over 9 must be equal to 1. Or x would be plus or minus 3. So the hyperbola is going to look like this. So this is at plus 3 comma 0. The hyperbola will open up like that. And then at 1, 2, 3, negative 3 comma 0, the hyperbola is going to open up to the left. And so in the problem when they describe the points A and B, they're probably talking about that point A and that point B. Now, let's think about what this question is asking us. Equation of a common tangent with positive slope-- so it has to have a positive slope-- to the circle as well as to the hyperbola-- a common tangent. So let's just think about this a little bit. So it's going to have a positive slope, so it won't be tangent to the circle anywhere where the circle has a negative slope. So it can't be tangent over here. It can't be tangent over there. And then we could say, well, if it was tangent to the circle over here, what would happen? Well , it wouldn't be able to be tangent to the hyperbola. So it has to be tangent to the circle someplace in this blue region right over here. And then how can it be tangent to the hyperbola? It might be tempting to say that it would be tangent to the hyperbola in this way somehow, but what you need to realize is the hyperbola is asymptoting towards some line. And we could figure out what that line is. Its asymptoting towards some line. So let me draw that line. The hyperbola is always going to have a higher slope than that line, a very slightly higher slope. It's slowly approaching that line. So if you go out here, the hyperbola is going to have a higher slope than the asymptote line. And so if you had to be tangent to it, you would have to have a higher slope. And anything that's coming from this part of the circle towards anything out here on the hyperbola is going to have to have a lower slope than the tangent line, right? Because it's going to have to catch up. The tangent line is going to have to catch up to whatever-- let me draw it again-- to whatever we draw over here. I want to make this clear. The hyperbola, as you go out here, actually this whole period, this whole part of the hyperbola, is going to have a higher slope than what it is asymptoting towards. That's what allows it to get closer and closer to that line. So any tangent is going to have to have a higher slope out here. Anything tangent to the hyperbola is going to have to have a higher slope than this actual line. It's going to have to have a slightly higher slope. So if we take something out here and we try to draw a tangent from this part of the circle out there, it wouldn't work. Because this tangent, by definition, in order to meet the hyperbola, is going to have to have a lower slope than this asymptote. So this can't be tangent to that part of the hyperbola. So what else can we do? Well, the only other part of the hyperbola that we might be able to work something out is this part of the hyperbola right over here. So if we find a line that's tangent maybe there and there on the hyperbola, then we might have found our common tangent with positive slope. So let me draw that. So our common tangent with positive slope, I'll do that in pink. Our common tangent with positive slope could look like that. So now that we have the visualization down, in the next video let's try to figure out what that line might look like, especially when we constrain it to having to be tangent to the circle and having to be tangent to this hyperbola.