# more circles!!

Posers and Puzzles 25 Oct '04 17:32
1. 25 Oct '04 17:32
circle w/ r=2 centered @ (2,1)

the line x=0 is tangent.

find an equation for the other line tangent to this circle passing through (0,0)
2. telerion
True X X Xian
25 Oct '04 20:27
circle w/ r=2 centered @ (2,1)

the line x=0 is tangent.

find an equation for the other line tangent to this circle passing through (0,0)
Should this be centered at (2,2)?
3. AThousandYoung
All My Soldiers...
25 Oct '04 20:35
No, that would be easy. The problem as given is much more challenging.
4. telerion
True X X Xian
25 Oct '04 20:41
Originally posted by AThousandYoung
No, that would be easy. The problem as given is much more challenging.
Forget it, dumb me. x=0 is the y-axis. Not thinking; damn midterm this morning through me all out of kilter.
5. telerion
True X X Xian
25 Oct '04 21:452 edits
Ok maybe I'll screw up the math here but oh well.

So we must find an equation for a line.

The form of the line is y=mx+b where b=0.

Now it should be obvious to the most casual observer that the point of tangency must be on the lower half of the circle.

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

(y-1)^2 = 4 - (x-2)^2

So the lower half is

(1)y =-sqrt(4-(x-2)^2)+1

Now in order for our line to be tangent to the circle two conditions must hold.

First the line and the circle must intersect, that is

(2) mx= -sqrt(4-(x-2)^2)+1

And second, the slope of the line, m, must be equal to the slope of the of the curve -sqrt(4-(x-2)^2)+1 at that point of intersection. This will give us a system of two equations in two unknowns.

Now taking the first derivative:

(3) m=(x-2)/(sqrt(4-(x-2)^2))

Now substitute back into (2)

(4) x*(x-2)/(sqrt(4-(x-2)^2))=-sqrt(4-(x-2)^2)+1

This yields x=4/5.

Now plug this x back into (3)
m=-3/4

So the equation of the line is

(5) y= -(3/4)*x

Any glaring errors?

Oh yeah and the point of intersection is (x,y)=(4/5,-3/5)
6. 26 Oct '04 20:321 edit
[i]2)

(4) x*(x-2)/(sqrt(4-(x-2)^2))=-sqrt(4-(x-2)^2)+1

This yields x=4/5.

Now plug this x back into (3)
m=-3/4

So the equation of the line is

(5) y= -(3/4)*x

Any glaring errors?

Oh yeah and the point of intersection is (x,y)=(4/5,-3/5)[/b]
no, that looks right. this was in my calc homework, it took my class fiften munutes to get to the first tangent, i dont think they would have got the other if i hadn't waken up. (BragðŸ˜€BragðŸ˜€Barg) we did it a more geometric way, rather than algebreic. i think your way is better.
7. Palynka
Upward Spiral
27 Oct '04 09:02
I think I'm probably making more steps than necessary, but:

The line is perpendicular with the radius (as tangent) therefore there is a rectangular triangle with (0,0), (2,1) and the tangency point (a,b).

We know the length of the hypotenuse (sqrt (5) ) and one of the legs (2-radius). Therefore the distance of this point to (0,0) is 1 and the distance to (2,1). is 2.

1 = sqrt( (a-2)^2 + (b-1)^2)
2 = sqrt( (a-0)^2 + (b-0)^2)

Now it's easy, two equations and two variables. (note that we know a is positive and b is negative as the point has to be in the lower right quadrant).

a=0.8 and b=-0.6

And the following equation for the line that goes through this point and (0,0) is:

y = -0.75*x

8. telerion
True X X Xian
27 Oct '04 12:46
Originally posted by Palynka
I think I'm probably making more steps than necessary, but:

The line is perpendicular with the radius (as tangent) therefore there is a rectangular triangle with (0,0), (2,1) and the tangency point (a,b).

We know the length of the hypotenuse (sqrt (5) ) and one of the legs (2-radius). Therefore the distance of this point to (0,0) is 1 and the distance ...[text shortened]... ollowing equation for the line that goes through this point and (0,0) is:

y = -0.75*x

Cool!