Tangent Line

[Raydain]

I'm taking a calculus pre-test but I forgot how to calculate tangent lines to a point.

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

Off the top of my head I'd guess to calculate tangent lines to the circle in the general form:

y = m*x + b(m)

where b(m) is your intercept as a function of m

I'm guessing there will be two general equations here, one for m>0 and one for m<0

then plug in your point to find m in each equation

watch out for m=0 and m=infinity though, as this method won't account for these cases

edit: This may or may not be correct, and may or may not be the easiest way

[udontknowmickey]

I think Hotpants is right, but to clarify it a little

the line will be of the form

Ax+By+C=0 (I prefer general linear)

it intersects the circle x^2+(y+1)^2=1 at one point
it contains the point (0,1)

Thus we have two eqns:
from the point (0,1) we have:

A(0)+B(1)+C=0
B=-C
so
Ax+By-B=0
and
x^2+(y+1)^2=1
has one solution (x,y), we're looking for A,B.

well, we can solve the line for y and plug into to solve the circle:

y=1-A/Bx and we get:

x^2+(1-A/Bx+1)^2=1

and you have a quadratic. If it intersects at only one point, then that means the solution for x is when the descriminant=0. Plug this in and you should have a quadratic solving A in terms of B. Do that and you'll get the eqn.

That is hard. Here is a much easier way:

Notice that the tangent line will be tangent to the radius of the circle. Thus for each point (x,y) on the circle, the slope of radius to that point is computed as (y+1)/x. If the slope of the line is tangent to this radius, then the slope of the line is the negative inverse at that point: -x/(1+y)

Thus we have a simple(r) system of eqns.

line contains point (x,y) and (0,1) resulting in a slope of (y-1)/x which is the same as -x/(1+y) and so we get (simplifying) x^2+y^2=1. This is the locus of points where the slope of the line to that point is equal to our expressions. Now we just need where that collection of points intersect our circle: x^2+(y+1)^2=1

Subtract the two eqns and we quickly have 2y+1 =0 y= -1/2
leading to x=+/- rt(3)/2.

Hope this doesn't come too late. Good luck with doing it the method the book prescribes

[Darryl_P]

I think you meant "perpendicular to the radius" and not "tangent to the radius".

Good analysis, though. I would have responded to the OP but it seems you have more or less nailed it.