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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition) > Unit 2

Lesson 12: Challenging conic section problems (IIT JEE)- Representing a line tangent to a hyperbola
- Common tangent of circle & hyperbola (1 of 5)
- Common tangent of circle & hyperbola (2 of 5)
- Common tangent of circle & hyperbola (3 of 5)
- Common tangent of circle & hyperbola (4 of 5)
- Common tangent of circle & hyperbola (5 of 5)
- Intersection of circle & hyperbola

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# Intersection of circle & hyperbola

2010 IIT JEE Paper 1 Problem 46: Find the equation of a circle whose diameter is defined by the two intersection points of given circle and hyperbola. Created by Sal Khan.

## Want to join the conversation?

- I must have missed it somewhere, why we need to complete the square at the start? for the circle equation? whats the reason behind it?(3 votes)
- It's to put the equation in the standard form to make the radius clear.(6 votes)

- at0:45Mr.Khan adds 16 to the left but adds 6 to the left

and my math teacher says whatever you do to the left you have to do to the right...?

or nawwwww(0 votes)- That was most certainly a written error. It doesn't affect the rest of the question because he continued the addition correctly.(11 votes)

- At9:22, Sal dismissed a negative extraneous root of x. How was that extraneous root created? I was curious because I did not see Sal square both sides of an equation or do anything else that would produce a false root.(4 votes)
- he does not need the negative root

the intersection is on right side of y axis(2 votes)

- fast solution: first, look at the choices, then graph both of them, look at the intersection points, predict the y and x from looking at it, and then substitute for the equation of the circle. expand... you choose the answer closest to yours.(4 votes)
- How did Sal get the 4 points that made the circle in the first part of the video?(2 votes)
- He used the equation of the circle that was already given.(4 votes)

- How do we know that the center was (6,0)?. I understood the equation that he did was for the points in the circle not for the center. Help please!(1 vote)
- The midpoint of the diameter (center of circle) with end points (6, root 12) and (6, - root 12) would give (6, 0)(2 votes)

- but we just found the positive slope of the hyperbola not the negative one.so at3:25how would sal prove the other line from the equation of the hyperbola.when he seid " And it's going to be symmetric ".(1 vote)
- what if x is not a pretty number like 15plus or minus square root of 305 divided by 10?(1 vote)
- They are probably not going to give a question like this which does not result in relatively simple values of x.(1 vote)

- How do you know to Solve for y^2? I tried making the equation of the circle and the hyperbola equal to 1 and then setting them equivalent, which got me to a dead end at 7x^2 + 72x - 45y^2 = 0. I understand the tools we're being taught, but I'm struggling with how to strategically solve a problem. I'm just blindly algebraically manipulating things and going in circles or dead-ending.(1 vote)
- This is OT but I have a problem anwering the questions about rectangular forms of complex numbers when i write : 14sqrt[cos 135° + i sin 135°] the program tells me that it doesn't understand I used ALT+248, ALT 0176 and ALT 0186 to make the "degree" sign.

But the program still doesn't understand.

What am I doing wrong?(1 vote)- Don't use the degree sign. Just put it in manually like cos(135). If you need to, convert the degrees to radians first.(1 vote)

## Video transcript

The circle x squared plus
y squared minus 8x is equal to 0, and the
hyperbola x squared over 0 minus y squared
over 4 is equal to 1, intersect at the points
A and B. In problem 46, they want us to find equation
of the circle with AB as its diameter. So let's visualize the
circle and the hyperbola. The equation of the
circle x squared plus y squared minus 8x. Let me write it this way. This can be rewritten as x
squared minus 8x plus y squared is equal to 0. You can add 16 to both sides. And I'm doing that to complete
the square for the x term. So plus 16. This over here becomes
x minus 4 squared, plus y squared is equal to 16. And so the circle's going
to look something like this. That's my x-axis. That's my y-axis. Its center is at 4, 0. Zero. So 1, 2, 3, 4-- 4, 0. And its radius is 4. Its radius squared is 16. So its radius is 4. So let me go 1, 2,
3, 4-- 1, 2, 3, 4. Then we go up 4-- 1, 2, 3, 4. So all of these points, that
point, that point, that point, and that point are all
going to be on the circle. Let me draw that a
little bit neater. That point, that
point, and that point are all going to
be on our circle. So my circle's going to
look something like this. It's not drawn as
neatly as it could be. But I think you get
the general idea. This is our circle. Then they have a hyperbola. x squared over 9 minus y squared
over 4 is going to equal 1. This hyperbola is going to
open to the left and the right since our x term is positive. It's going to look
something like this. You can actually figure
out the asymptotes. Actually, let me just do that. Let me solve for y. So you would get, let
me write it this way, negative y squared over 4 is
equal to negative x squared over 9 plus 1. I just subtracted x squared
over 9 from both sides. And then we would get,
let's multiply both sides by negative 4. You get y squared
is equal to 4/9. X squared minus 4. Or y is equal to--
and now this will just be the positive part over here. But I'm doing that so that we
can understand its asymptotes. y is equal to the square root
of 4 over 9x squared minus 4. So as x gets larger
and larger and larger, this term right here is going
to stop mattering so much. So as x approaches
infinity, this is going to approach the
square root of 4/9 x squared. This constant's not
going to matter much. And so it's going to approach,
this thing right over here is going to be 2/3 x. So it's going to, if you
imagine a slope, let's see, 2/3. If you run 3, you rise 2. So the asymptote is
going to look like that. It's going to approach
that line over there. And it's going to be symmetric. So it's also going to
approach-- so 3 and 2, it's also going to approach
that line over there. And if we want to see where
it intersects the x-axis, you just set y is equal to 0. You get x squared
over 9 is equal to 1. So x is going to be
equal to plus or minus 3. So the positive intersection
is going to be over there. So our hyperbola
on the right side's going to look something like,
something, something like that. It's also going to show
up on the left side. But that's less interesting
because it's not doing anything with the circle. Now they told us that the
hyperbola and the circle intersect at the points A
and B. So this right here is the point A. This
right here is the point B. In this question, we want the
equation of the circle with AB as its diameter. So AB as the diameter. I could have drawn that
a little bit straighter. So the equation of that
circle right over there. So essentially, we
just need to figure out where does this hyperbola
intersect this circle. Now, the easiest way to
do this is to-- well, we have two constraints here. Let's solve for y
squared for the circle. And we could substitute that
in for y squared right here and see what x values,
the intersect that. And then we want the
x value out here. So something that
looks like this value is the one we want to use. Then we could figure
out the y value. The y value is going to be
the radius of our circle. The x value comma 0
will be the center. And then we'll
have our equation. So let's do that. So this up here, we have--
let me do it in yellow. So if we subtract x squared and
negative 8x from both sides, the equation of
the circle can be rewritten as y squared
is equal to-- I'm going to add 8x to both sides. So it's 8x. And then subtract x
squared from both sides. 8x minus x squared. I really just moved
to that and that over to the right hand
side of the equation. Now I can take
this and substitute for y squared in the
equation of the hyperbola. So the hyperbola's
equation is x squared over 9 minus y squared
over 4 is equal to 1. Instead of running a
y squared over there, we know that it has to
satisfy this equation as well. So for y squared, I'm going
to put an 8x minus x squared. And let's see if we can
solve this right here. This is just a straight
up quadratic equation, although it might not
look like it just yet. Let's simplify this. So this tells us that x
squared over 9 minus 8x/4. So minus 2x plus x squared
over 4 plus x squared over 4. I just distributed
essentially the negative 1/4 over both these terms. Is equal to 1. Let's see, we can multiply
the whole thing times 36 to get rid of these fractions. So that's 4 times 9. So 36 divided by 9 is 4. So this is 4x squared
minus 36 times 2 is 72. 72x plus 36 divided by 4 is 9,
plus 9x squared is equal to 36. These two terms right
here, we can add them. We get 13x squared, minus 72x. And then we could subtract
36 from both sides. So minus 36 is equal to 0. So now we just have a straight
up quadratic equation. We just have to find
the x's, find the roots. So let's use the
quadratic formula here. So x is going to be
equal to negative B So that's negative negative 72. So it's 72 plus or minus the
square root of 72 squared. I'll just write that 72
times 72 minus 4 times A times C. C is negative 36. So that negative, you
could put it out here. And just put the 36 out back. And then all of
that over 2 times a. So all of that over 26. So the hard part is
really to simplify this. But it looks like we have some
interesting stuff going on. So let me rewrite
this part over here. Let me write it over here. 72 times 72 is the same thing
as 2 times 36 times 2 times 36, right? Each of those are 72. That's the same
thing as 72 times 72. And then we're adding to
that 4 times 36 times 13. And we're taking the square
root of this whole thing. I'm just doing the work
out here so we don't. waste this real
estate over here. Now we can factor
out a 4 times 36. This is 4 and this is a 36. So this is equal
to-- I know it's getting-- let me write
it a little bit neater. This is equal to the square root
of, we can take a 4 times 36. And actually 4 times 36 is 144. I'll just write 4
times 36 right here. 4 times 36. Now over here we use
the both 2s in the 36. So we have a 36 left. We use this 4 times 36. So we have plus 13. So this becomes the square
root of, 4 times 36 is 144. 36 plus 13 is 49. So we're lucky that
it actually worked out to 2 perfect squares. So this is the square root
of 144 times the square root of 49, which is 12, times
7 or this is equal to 84. So this business over here
simplifies to 72 plus or minus 84. This whole thing over
here is simplified to 84, all of that over 26. Now if we were to
subtract 84, we would get something
over here that doesn't make sense
in this context. We're looking for something
a positive x value. So let's only consider the
plus 84 and see what we get. So the first thing to
do, actually let me just divide the numerator and
the denominator by 2. So this is the same thing
as 36 plus 42 over 13. And this is the same
thing as 78 over 13. And it looks like 78
is divisible by 13. 78 divided by 13 is 6. This is equal to 6. So the x-coordinate here is 6. It's right there. So it's 6. We don't know what
the y-coordinate is. So let's do that. It's pretty easy to solve. We could substitute x equals
6 into any of these equations. This one is probably
easier right over here. So we get y squared
is equal to 8 times 6 is 48, minus 48 minus
36, which is equal to 12. And so y is equal to
the square root of 12. We could simplify that
radical if we want. But we know the point now. It's 6 comma the square root--
Sorry, this point over here is 6 comma 0, but the
actual intersect, the center is 6 comma 0, but the
actual intersection point is going to be 6 comma
positive square root of 12? And I should say this is
a positive or negative square root of 12. And this row over
here is going to be 6 negative square root of 12. This right here is 6 comma 0. That's the center of our circle. So what's the equation of
this new circle going to be? What's the equation of this
new circle going to be? Well we know its center. Its center is at 6 0. Let me write it down here. Its center is at 6 0. So it's going to be x minus 6
squared plus y minus 0 squared is equal to the radius squared. Now what's the radius? The radius is this
height right over here. Or it's just equal to the y
value at the intersection. It's the square root of 12. Now this is going to
the radius squared. The radius is squared of 12. Squared of 12 squared is 12. So the equation is
going to be x minus 6, and see what form
they have up here. So they actually
multiply everything out. So let's just do that. So this is going to be x
squared minus 12x plus 36 plus y squared is equal to 12. And then we could subtract
12 from both sides. Subtract 12 from both sides. And we get x squared minus
12x plus 24 plus y squared is equal to 0. Let's see which of the
choices match that. Let me copy it. And then let me
paste it up here. Let me paste it up here. And it looks like we have
an x squared positive y squared, a negative 12x
x and then a positive 24. Negative 12 x positive 24. So it looks like
our answer is a. Did I do that right?
x squared plus y squared minus 12 x plus 24. Yep, our answer is a.