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## Challenging conic section problems (IIT JEE)

Current time:0:00Total duration:6:16

# 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.