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

• JOhnJDC
In summary, John tried to solve for the equations of the tangent lines to the circle using the equation of the circle and the slope of the tangent line, but wasn't able to solve it. He found the slope of the tangent line by drawing a radius from the center of the circle to the point of tangency and solving for the slope. From there, he was able to find the equation of the line through the points (0,4) and (a,b) and found that it has slope m = (b-4)/a. The line through the points (0,1) and (a,b) has slope n = (b-1)/a and since they are perpendicular, their slopes are
JOhnJDC
Hello - I've been stuck on this for a while now and I really need some help.

## Homework Statement

Find the equations for all lines that are tangent to the circle x^2+y^2=2y and pass through the point (0,4).

y=mx+b
ax^2+bx+c=0

## 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

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.

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.

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

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

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):

$$\frac{b-4}{a} = \frac{a}{1-b}$$

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

$$a^2 = (b-4)(1-b)$$

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

So now we have two different expressions for a^2:
$$$a^2 = (1-b)(b-4)$ $a^2 = 1 - (b-1)^2$$$

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

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
$$a = \pm \frac{2\sqrt{2}}{3}$$

Then you can plug these into one of your expressions for m, and you'll get something like
\begin{align*} m &= \frac{a}{1-b}\\ &= \frac{\pm\frac{2\sqrt{2}}{3}}{1 - \frac{4}{3}}\\ &= \mp 2\sqrt{2} \end{align*}
("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
$$y = \pm2\sqrt{2}x + 4$$

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

John

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..

## 1. How do I find the equation of a tangent line to a given circle from an external point?

To find the equation of a tangent line to a given circle from an external point, you can use the point-slope form of a line and the equation of a circle. First, determine the coordinates of the external point and the center of the circle. Then, use the Pythagorean theorem to find the length of the radius of the circle. Finally, plug the values into the point-slope form to find the equation of the tangent line.

## 2. What is the point-slope form of a line?

The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line. This form is useful for finding the equation of a line when given a point and the slope.

## 3. Can you provide an example of finding the equation of a tangent line to a circle from an external point?

Sure, let's say we have a circle with center (2,3) and radius 5. The external point is given as (6,2). First, we use the Pythagorean theorem to find the length of the radius, which is √(62 + 12) = √37. Then, we plug the values into the point-slope form: y - 2 = (1/3)(x - 6). This simplifies to y = (1/3)x + (4/3), which is the equation of the tangent line.

## 4. What if the external point is inside the circle?

If the external point is inside the circle, there will be no tangent line because the point will be part of the circle. In this case, the equation of the tangent line is undefined.

## 5. Is there a shortcut for finding the equation of a tangent line to a circle from an external point?

Yes, there is a shortcut called the "power of a point" theorem. This states that the product of the distances from an external point to the tangent points on a circle is equal to the square of the length of the tangent line. So, you can use this theorem to find the equation of the tangent line without having to use the Pythagorean theorem.

• Precalculus Mathematics Homework Help
Replies
18
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
703
• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Precalculus Mathematics Homework Help
Replies
19
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
817
• Precalculus Mathematics Homework Help
Replies
2
Views
586
• Precalculus Mathematics Homework Help
Replies
2
Views
977
• Precalculus Mathematics Homework Help
Replies
8
Views
2K
• General Math
Replies
2
Views
545
• Precalculus Mathematics Homework Help
Replies
17
Views
1K