NationStates Jolt Archive

Anyone? Anyone?

What's the problem?

What's Cal?

Well, there are three...

One of them is this...

There are two tangent lines from the point (0,1) to the circle x^2+(y+1)^2=1. Find equations of these two lines by using the fact that each tangent line intersects the circle in exactly one point.

I know that one equation will be y=mx+b and the other y=-mx+b, and I know that b=1, but I need to figure out what m is equal to.

Oh, by the way, I don't want the answer. I want to know how to work the problem.

What's Cal?
Calculus.

Calculus.

Thanks, I'm British.

I also know that the line from the center of the circle to the tangent point is 1, the line from the center of the circle to the intersection point is 2, and from the pythagreon theorem, I know the distance from the intersection point to the tangent point is the square root of 3.

No one here knows any math, eh?

I do beleive you need to determine the derivative of the circle function at the tangent(s) points to find slope m of each linear function.

It's been a long time though.

*scratches head*

take the derivative of the equation of your circle, once with respect to x, and again with respect to y. this gives you tangent points.... from there, you're right with the y=mx + b.

and don't forget that when you take the square root of an integer, you have a positive answer and a negative answer.

oops I should say circle formula ..circles are not functions.

take the derivative of the equation of your circle, once with respect to x, and again with respect to y. this gives you tangent points.... from there, you're right with the y=mx + b.
...so...what does that mean in English (this is my first ever Cal class...and the teacher only speaks math and the students only speak English)

No one has a Math-to-English translator?

No one has a Math-to-English translator?
Nope, no one.

x^2+(y+1)^2=1

and

y=mx+1

...so....

x^2+(mx+1+1)^2=1
X^2+(mx)^2+4mx+4=1

am I on the right track? I just need to solve for m?

Crap, my dad just left. He's a calculus teacher, he could have helped you out no problem.

Meh...I could probably use trig to figure out the tangent point then solve for the slope based of that point and the point (0,1)...but my Cal teacher might not like that...

Well, there are three...

One of them is this...

There are two tangent lines from the point (0,1) to the circle x^2+(y+1)^2=1. Find equations of these two lines by using the fact that each tangent line intersects the circle in exactly one point.

The first line is y=(sqrt3)x+1, and the other is y=-(sqrt3)x+1.

how is this calculus?

just one year removed from college and already this is making my head hurt...

Er, I know you don't actually need calculus to find the tangent line to a circle. So you won't need to mess around with derivatives or anything like that. (That comes later, poor unsuspecting one...)

Er, I know you don't actually need calculus to find the tangent line to a circle. So you won't need to mess around with derivatives or anything like that. (That comes later, poor unsuspecting one...)

what, you don't like derivatives? that's childs play - give me continuum mechanics in tensor notation....now that's challenging

Well, there are three...

One of them is this...

There are two tangent lines from the point (0,1) to the circle x^2+(y+1)^2=1. Find equations of these two lines by using the fact that each tangent line intersects the circle in exactly one point.

Here's how you do it:

First, you take the derivative of y with respect to x, y':

x^2+(y+1)^2=1
2x+2(y+1)y'=0
x+(y+1)y'=0
(y+1)y'=-x
y'=-x/(y+1)

Ok, so that tells you an expression for the slope of ANY line tangent to the circle.

Now, you want the two tangents that go through the point (0,1). Well, since these lines go through the circle, their slopes at the points of intersection are both -x/(y+1). But then

y=[-x/(y+1)]x+1
y=-(x^2)/(y+1)+1
(y-1)=-(x^2)/(y+1)
(y-1)(y+1)=-x^2
y^2-1=-x^2
y^2+x^2=1

This is a circle, which may look like we've gone backwards! But in fact, now we know that the points we are looking for satisfy the TWO equations:

x^2+y^2=1
x^2+(y+1)^2=1

Subtracting equation 2 from equation 1 gives:

y^2-(y+1)^2=0
y^2-(y^2+2y+1)=0
y^2-y^2-2y-1=0
-2y-1=0
-2y=1
y=-(1/2)

Ok, so you know y=-(1/2), which should make perfect sense if you've drawn a graph of the circle and its tangents. Plugging -(1/2) in for y in the original equation gives:

x^2+[-(1/2)+1]^2=1
x^2+(1/2)^2=1
x^2+1/4=1
x^2=3/4

x=(sqrt3)/2
x=-(sqrt3)/2

So your tangents intersect the circle at (-(sqrt3)/2, -1/2) and ((sqrt3)/2, -1/2). Finding the slopes is easy:

1. (-1/2-1)/[-(sqrt3)/2]
-(3/2)/[-(sqrt3)/2]
-3/-(sqrt3)
3/(sqrt3)
sqrt3

(-1/2-1)/[(sqrt3)/2]
-(3/2)/[(sqrt3)/2]
-3/(sqrt3)
-sqrt3

Edit: You can also get the slopes by plugging those points into the equation for the derivative of the circle, y'=-x/(y+1):

1. y'=-(sqrt3 / 2)/(-1/2+1)
-(sqrt3 / 2)/(1/2)
-sqrt3/1
-sqrt3

2. y'=-(-sqrt3 / 2)/(-1/2+1)
(sqrt3/ 2)/(1/2)
sqrt3/1
sqrt3

Thus, the lines I mentioned:

1. y=(sqrt3)x+1
2. y=-(sqrt3)x+1

If you don't believe it, just plug the points (-(sqrt3)/2, -1/2) and ((sqrt3)/2, -1/2) into those two equations and see that they work... and then into the circle, to see that they work there, too.

Wow, I knew this about three months ago when I was still in school... But it seems like other people are on the right track. All so familiar, yet makes my head hurt all the same. Good luck!