more circles!!

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)

Originally posted by fearlessleader
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)?

No, that would be easy. The problem as given is much more challenging.

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.

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.

To get this equation for this semi-circle start with
(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)

[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. (Brag😀Brag😀Barg) we did it a more geometric way, rather than algebreic. i think your way is better.

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

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!