Equation of a tangent line to a given circle from an external point - Stuck!


by JOhnJDC
Tags: circle, equation, point, quadratic, tangent
JOhnJDC
JOhnJDC is offline
#1
Nov28-09, 06:07 PM
P: 36
Hello - I've been stuck on this for a while now and I really need some help.

1. The problem statement, all variables and given/known data
Find the equations for all lines that are tangent to the circle x^2+y^2=2y and pass through the point (0,4).


2. Relevant equations
y=mx+b
ax^2+bx+c=0


3. The attempt at a solution
From the given equation of the circle, I know that the circle has a center of (0, 1) and that its radius is 1. I also know that the general equation of the tangent line is y-4=m(x-0) or y=m(x-0)+4, which really is simply y=mx+4, right?. I'm trying to solve for the equations of the tangent lines by plugging y=mx+4 into the equation of the circle, x^2+y^2=2y, and solving the resulting quadratic equation. However, I can't seem to solve it. Is this a correct approach to this problem? Here is what I did:

x^2+(mx+4)^2=2(mx+4); simplifying:
x^2+(m^2)(x^2)+6mx+8=0

Assuming this is the right approach so far, I'm not really sure what to do from here. I applied the quadratic formula to the "mx" part of the equation ((m^2)(x^2)+6mx+8) and got -2 and -4, but I think I'm missing something. I'd really appreciate a walk-through.

Many thanks,
John
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Mark44
Mark44 is offline
#2
Nov28-09, 08:38 PM
Mentor
P: 21,012
I think that what you're missing is that the tangent line is tangent to the circle. Since you posted this in the Precalc section, I'm assuming you don't know calculus, but it's not needed in this problem.

You have the equation of your tangent lines (there are two of them) as y = mx + 4, which has a slope of m. Draw a radius from the center of the circle to the point of tangency (x, y). Can you find the slope of this radius knowing that it is perpendicular to the tangent line? Also, the point of tangency is on the circle, so it has to satisfy the circle's equation.
xeno_gear
xeno_gear is offline
#3
Nov28-09, 08:39 PM
P: 40
Sorry, accidentally hit submit before I was finished. Anyway..

Your attempt is good, but doesn't use the fact that the lines are tangent to the circle.
First you'll want the center of the circle.

[tex]
\begin{align*}
x^2 + y^2 &= 2y\\
x^2 + y^2 - 2y &= 0
\end{align*}
[/tex]

Complete the square (I'll leave the details to you) to get:
[tex]
x^2 + (y-1)^2 = 1
[/tex]

So the center of the circle is (0, 1), and the radius is 1.

You're right about the equation of the tangent line. But what you also need to use is that the tangent line to a circle at point P is perpendicular to the line that goes through point P and the center of the circle.

So you want to find the equation of all lines that go through (0, 4) and are tangent to the circle. Such a line will hit the circle at some point (a, b). So you want a line that goes through the points (0,4) and (a,b). This line is perpendicular to the line that goes through (a, b) and (0, 1), since (0, 1) is the center of the circle.

The line through the points (0,4) and (a,b) has slope m = (b-4)/a.
(That's the line you want to find.)

The line through the points (0,1) and (a,b) has slope n = (b-1)/a.

Since these lines are perpendicular, their slopes are negative reciprocals. So you have
m = -1/n, which means (after some simplifications):

[tex]
\frac{b-4}{a} = \frac{a}{1-b}
[/tex]

Multiply both sides by a*(1-b), and you get

[tex]
a^2 = (b-4)(1-b)
[/tex]

Note that the point (a,b) is on the circle, so it satisfies the equation of the circle. In other words, we have
[tex]
\begin{align*}
a^2 + (b-1)^2 &= 1\\
a^2 &= 1 - (b-1)^2
\end{align*}
[/tex]

So now we have two different expressions for a^2:
[tex]
\[a^2 = (1-b)(b-4)\]
\[a^2 = 1 - (b-1)^2\]
[/tex]

So you want to set them equal to each other and solve for b:
[tex]
\begin{align*}
(1-b)(b-4) &= 1 - (b-1)^2\\
b - 4 - b^2 + 4b &= 1 - (b^2 - 2b + 1)\\
&\vdots
\end{align*}
[/tex]

I'll leave the details to you, but you should end up with b = 4/3. Then you plug that back into either one of your equations for a^2, and you'll get
[tex]
a = \pm \frac{2\sqrt{2}}{3}
[/tex]

Then you can plug these into one of your expressions for m, and you'll get something like
[tex]
\begin{align*}
m &= \frac{a}{1-b}\\
&= \frac{\pm\frac{2\sqrt{2}}{3}}{1 - \frac{4}{3}}\\
&= \mp 2\sqrt{2}
\end{align*}
[/tex]
("m" has the oppposite sign of "a".. that is, if one is positive, the other is negative)

So the equations of your tangent lines are
[tex]
y = \pm2\sqrt{2}x + 4
[/tex]

JOhnJDC
JOhnJDC is offline
#4
Nov28-09, 09:39 PM
P: 36

Equation of a tangent line to a given circle from an external point - Stuck!


Thanks so much, xeno_gear. I followed your logic entirely and arrived at the correct answer.

John
xeno_gear
xeno_gear is offline
#5
Nov28-09, 09:42 PM
P: 40
no problem. haha, yeah, i did that at first, then i graphed it to check and realized there was a problem. i did edit it, but for some reason it didn't update right away, i dunno..


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