If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Precalculus (2018 edition)

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

Lesson 12: Challenging conic section problems (IIT JEE)

# Common tangent of circle & hyperbola (5 of 5)

2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 5. Created by Sal Khan.

## Want to join the conversation?

• There must be a very shorter way to solve the question unlike this solution. Can you make a tutorial on shorter ways to solve this problem ? Because we definitely are given 3 hours to solve the whole question paper.
• You can find the tangent slope vs. y-intercept equation for both the hyperbola and circle pretty easily, a minute each maybe, then all you need to do is plug in the slope and y-intercepts of the different answer choices until they're equal. Sal made this way harder
• Wow... what is the time limit in the JEE exam to solve this problem?
• 2.5 min MAXIMUM!
• Is there a faster way to do this problem using calculus?
(1 vote)
• Joseph,

There is a faster way, but it does not require any calculus knowledge.

In general, when you solve problems under time constraints and you are provided with answer choices, one potential technique is to check whether each answer choice is a solution to the problem at hand. According to this approach, you should not try to find general solution to the problem and check which answer choice corresponds to your findings (what Sal did for this question), but rather try to go the other way around: substitute each answer choice to your equations and check which answers satisfy the constraints of the problem. This technique may often help you to find correct answers faster.

Let us apply this approach to our issue. First, we need to ensure that we fully understand the situation and the objective. In this case we do so by plotting the circle and the hyperbola as Sal did at the beginning of the corresponding video: https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-1.

It is clear that we are looking for a line that is tangent to both the circle and the hyperbola. This means that the line should have exactly one common point with both the circle and the hyperbola.

After understanding the question, let us look at our answer choices. We need to essentially find an equation of a tangent, which is a line. Therefore, we would like to rewrite each of the answer choices in a standard form, so that we could quickly analyze them:
`(A) y = (2⁄√5)⋅x − 20⁄√5`
`(B) y = (2⁄√5)⋅x + 4⁄√5`
`(C) y = (3⁄4)⋅x + 2`
`(D) y = (4⁄3)⋅x + 4⁄3`

By the logic explained at around in https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-2, the y-intercept of our tangent should be positive. If we look at our answer choices with this in mind, we can immediately recognize that choice (A) is not correct. It is the only answer that has a negative y-intercept of −20⁄√5. As our y-intercept should be positive, we can automatically eliminate choice A.

As you can see, by simply looking at the problem and slightly modifying the answer choices, we have already been able to eliminate one of the answers, raising our probability of guessing correctly from 25.00% to ≈ 33.33%.

The next step should be to check whether our 3 potential candidates for a tangent satisfy the first condition: they should have exactly one common point with our circle. A tangent is a line and so it has an equation of y = mx + b. Instead of finding m and b that satisfy our constraints (Sal has already done this for us), we will take pairs of m and b from our answer choices and check whether these pairs satisfy our constraints.

As an example, for choice (B): y = (2⁄√5)⋅x + 4⁄√5, m = 2⁄√5 and b = 4⁄√5.

To check whether the answer choices satisfy the first condition of having exactly one common point with our circle, we may simply substitute our y from the equation of a tangent y = mx + b to the equation of the circle (Sal explains it at around in https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-2):
`y = mx + b` – equation of the tangent
`x² + y² − 8x = 0` – equation of the circle
`x² + (mx + b)² − 8x = 0` – equation for x where the tangent intersects with the circle

In order for our tangent and our circle to have exactly one common point, this new equation
x² + (mx + b)² − 8x = 0 should have only one solution. This is a quadratic equation, hence what we really need to do to check if an answer choice satisfies the condition is to plug corresponding m and b from this answer choice to the equation above and to test whether the discriminant of the resulting equation is 0.

We can work through the choice (B) together, so that you can see what I am talking about. The first step is to substitute m and b from choice (B) into the equation of our first condition:
`x² + (mx + b)² − 8x = 0`
`x² + ((2⁄√5)⋅x + 4⁄√5)² − 8x = 0`

Simplifying:
`x² + ((2⁄√5)⋅x)² + 2⋅(2⁄√5)⋅x⋅(4⁄√5) + (4⁄√5)² − 8x = 0`
`x² + (4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5 − 8x = 0`

Multiplying both sides of the equation by 5:
`5x² + 4x² + 16x + 16 − 40x = 0`
`9x² − 24x + 16 = 0`

Checking the discriminant:
`D = 24² − 4⋅9⋅16 = (8⋅3)² − 4⋅3⋅3⋅8⋅2 = 8⋅3⋅8⋅3 − 8⋅3⋅8⋅3 = 0`

The discriminant of the simplified equation of our first condition is indeed 0, which implies that there is only one x where our tangent and our circle intersect. Therefore, choice (B) does satisfy the first condition and cannot be eliminated at this stage.

Note that we did not need to solve for x in the equation above, we just needed to ensure that there is only one x that satisfies our equation. To do this we just check whether the discriminant is equal to 0.

If you substitute two other relevant answer choices (C) and (D) into the equation of our first condition
(x² + (mx + b)² − 8x = 0), you will see that for both answer choices the discriminants are also equal to zero. Therefore, all three answers that are left after we eliminated choice (A) satisfy the “exactly one common point with the circle” condition, and neither can be eliminated at this stage. This is unfortunate, but we must not despair!:) We just need to remember that there is another condition to test.

As you remember, our tangent line should not only have exactly one common point with the circle, but should also have exactly one common point with our hyperbola, which is going to be our second condition.

We can simplify the equation of the hyperbola:
`x²⁄9 − y²⁄4 = 1`

Multiplying both sides of the equation by 36:
`4x² − 9y² = 36`

Now, let us substitute our y from the equation of a tangent y = mx + b to the equation of the hyperbola:
`y = mx + b` – equation of the tangent
`4x² − 9y² = 36` – equation of the hyperbola
`4x² − 9⋅(mx + b)² = 36` – equation for x where the tangent intersects with the hyperbola

At this stage, we just need to substitute three answer choices that are left into this equation and find for which of them the equation has exactly one root. A single root means that our hyperbola and our tangent have exactly one common point, which satisfies our second condition. Let us work through the process for choice (B) first:
`4x² − 9⋅(mx + b)² = 36`
`4x² − 9⋅((2⁄√5)⋅x + 4⁄√5)² = 36`
`4x² − 9⋅[((2⁄√5)⋅x)² + 2⋅(2⁄√5)⋅x⋅(4⁄√5) + (4⁄√5)²] = 36`
`4x² − 9⋅[(4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5] = 36`

Multiplying both sides of the equation by 5:
`20x² − 9⋅5⋅[(4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5] = 180`
`20x² − 9⋅[4x² + 16x + 16] = 180`
`20x² − 36x² − 144x − 144 = 180`

Simplifying and subtracting 180 from both sides of the equation:
`−16x² − 144x − 324 = 0`

Multiplying both sides of the equation by −1:
`16x² + 144x + 324 = 0`

Looking at the discriminant:
`D = 144² − 4⋅16⋅324 = (12⋅12)² − 4⋅4⋅4⋅9⋅36 = (12⋅12)² − 12⋅12⋅4⋅3⋅12 = (12⋅12)² − (12⋅12)² = 0`

The discriminant for choice (B) is indeed equal to 0 and, therefore, our hyperbola and the tangent represented in choice (B) intersect at exactly one point, which is what we are looking for.

We showed that the tangent described in choice (B) has exactly one point of intersection with both our circle and our hyperbola and has a positive slope. It satisfies all the conditions of our problem and, therefore, choice (B) is the right answer!

It is also worth working together through the second condition using the answer (C):
`4x² − 9⋅(mx + b)² = 36`
`4x² − 9⋅((3⁄4)⋅x+2)² = 36`
`4x² − 9⋅[((3⁄4)⋅x)² + 2⋅(3⁄4)⋅x⋅2 + 2²] = 36`
`4x² − 9⋅[(9⁄16)⋅x² + 3x + 4] = 36`
`4x² − (81⁄16)⋅x² − 27x − 36 = 36`
`(64⁄16)⋅x² − (81⁄16)⋅x² − 27x − 72 = 0`
`−(17⁄16)⋅x² − 27x − 72 = 0`
`(17⁄16)⋅x² + 27x + 72 = 0`

Looking at the discriminant:
`D = 27² − 4⋅(17⁄16)⋅72 = (3⋅9)² − 4⋅(17⁄16)⋅4⋅18 = (3⋅9)² − 17⋅18 ≠ 0`

The discriminant in this case is clearly not equal to zero as (3⋅9)² does not contain 17 as a factor, but 17⋅18 does. In fact, the discriminant is positive (if you are not certain about this, simply calculate it, you will get 423), which means that the line described in choice (C) has two points of intersection with our hyperbola. If a line has two points of intersection with a hyperbola it cannot be its tangent by definition. Choice (C) does not satisfy the second condition of our problem and, therefore, choice (C) is NOT the right answer.

You can go through the same process to show that choice (D) is NOT the right answer as well, because when we plug the equation of the line from choice (D) into 4x² − 9⋅(mx + b)² = 36, we eventually get to a quadratic equation with the negative discriminant.

Overall, by simply plugging answer choices into the conditions of this problem and doing relatively easy algebra, we managed to avoid all the hairy calculations that we encountered together with Sal when trying to get to a general solution. This approach may save you some precious time during the actual exam and help to avoid unnecessary mistakes.

Have fun!:)
• A better way to solve this question would be by brute force method. If you check the options, only option (b) is actually tangent to the circle. Option (a) is directly eliminated as it has a negative y-intercept - which isn't possible. So the answer is clearly option (b). But of course, knowing the actual way to solve is also important. :)
• Just saying, but at roughly in the video, those 10s were divisible by 2, and since there where two 10s and two 2s, Sal could've factored the whole thing by another 4 on both sides before adding the stuff in the square root and continuing factoring... Just saying...
• This question was crazy.
• When you see a test problem like this that is this horribly nasty, there is usually a quicker solution using underlying principles. Other comments on this video have mentioned better ways to solve this faster. :)
• Sounds like he was out of breath.
• I think there's a better way to calculate the square root, like 50176, the last three numbers are 176, we can use 176 to minus the first two numbers 50, 176-50=126, and 126 is divisible by 7, so we can divide by 7 directly, it's easier and faster to calculate.
• So there is a faster way to solve this problem. If you know your polar coordinates, then you know that the circle can be given by r=8cos(Θ)*. Moreover,
*dy/dx = (sin^2(Θ)-cos^2(Θ))/(2sin(Θ)cos(Θ))* from (http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx) Then using slope-point form of a line, we know the equation of the tangent line is:
*y = dy/dx(x-rcos(Θ))+rsin(Θ)*
We know *r=8cos(Θ)* and *dy/dx
, so replacing that with our equation and simplifying, we should obtain:
y = (sin^2(Θ)-cos^2(Θ))/(2sin(Θ)cos(Θ))(x-8cos^2(Θ))+8sin(Θ)cos(Θ)*
Yes, this is kinda messy, but it works out nicely. At the end you should get:
*y = -xcos(2Θ)/sin(2Θ)+4cos(2Θ)(1+cos(2Θ)/sin(2Θ)+4sin^2(Θ)/sin(2Θ)*
*= -xcos(2Θ)/sin(2Θ)+(4cos(2Θ)+4cos^2(2Θ)+4sin^2(Θ))/sin(2Θ)*
*= -xcos(2Θ)/sin(2Θ)+4(cos(2Θ)+1)/sin(2Θ)*
Since 2Θ is present in all the functions, we can replace it with Θ. Thus:
*y = -xcos(Θ)/sin(Θ)+4(cos(Θ)+1)/sin(Θ)*
Great, that's our tangent line to the circle! If you know your formulas, you should be able to derive that very quickly.

Now the tangency constraint of a hyperbola is *c^2=9m^2-4
, when c and m represents the y-intercept and slope of the tangent line, respectively. Sal Khan derived this from https://www.khanacademy.org/test-prep/iit-jee-subject/iit-jee/v/tangent-line-hyperbola-relationship-very-optional.

We just derived our tangent line to the circle, so we know our m and c! By substitution, we have:
*(4(cos(Θ)+1)/sin(Θ))^2 = 9(cos(Θ)/sin(Θ))^2 - 4*
16(cos(Θ)+1)^2/sin^2(Θ) = 9cos^2(Θ)/sin^2(Θ) - 4
We can multiply both sides by sin^2(Θ) to remove the denominators:
16(cos(Θ)+1)^2 = 9cos^2(Θ) - 4sin^2(Θ)*
Multiplying the left-hand side results in:
*16cos^2(Θ)+32cos(Θ)+16 = 9cos^2(Θ) - 4sin^2(Θ)*
We can add 4sin^2(Θ) to both sides and use the fact that sin^2(Θ)+cos^2(Θ)=1:
*12cos^2(Θ)+32cos(Θ)+20 = 9cos^2(Θ)*
Taking all the stuff on the right to the left produces the quadratic:
*0 = 3cos^2(Θ)+32cos(Θ)+20

Okay, so this is a quadratic, so we can use the quadratic formula to see that cos(Θ) has two solutions, of which only one is viable:
cos(Θ)=-2/3 OR cos(Θ)=-10 (which cannot be)*

If *cos(Θ)=-2/3
then sin(Θ)=sqrt(5)/3.

Now that we know the values of cos(Θ) and sin(Θ), recall the tangent line we derived earlier:
y = -xcos(Θ)/sin(Θ)+4(cos(Θ)+1)/sin(Θ)*

Substitution for values results in our answer!
*y=2x/sqrt(5)+4/sqrt(5)* which results in
*2x-sqrt(5)y+4=0

If you know your formulas then you should be able to solve this very quickly.

Thanks for reading, I know formatting here is not that great.