# How can I find the equation of tangent lines from a circle to point A?

1. May 22, 2008

### JBD2

Circle: (x-2)² + (y+1)² = 25
Point A: (11,8)
Therefore Center of Circle = (2,-1) and Radius = 5

Looks somewhat like:
O>

Basically there are two tangents coming from the edge of a circle with the above equation to a point (pt. A). I have figured out to make two right triangles by putting a line between the center point of a circle to pt. A. I then got the distance between the center point and point A using the distance formula. Distance = 12.7279. Using the Pythagoras rule I solved each right triangle to get that the length of each tangent is 11.7047. I need to get the points though that connect the radii and tangents to the circle, so that I can get the slope of each tangent and write the equation for each tangent line. I'm unsure of how to do this, so can someone please help? Thanks.

2. May 23, 2008

### Tedjn

An easier way to do the problem may be to consider angles. What is the angle between the x-axis and the line connecting the center of the circle with point A? What is the angle between that line and the tangent line you want? What is the total angle between the tangent line and the x-axis? What is m = y/x in terms of that angle?

3. May 23, 2008

### JBD2

I'm not quite sure how I would find that angle though, because how would I know that the radius lines up with the x-axis? I know using sin cos or tan I could figure something out but I'm kind of unsure what to do. Can you show me what you mean?

4. May 23, 2008

### Tedjn

The radius doesn't have to line up with the x-axis. You can move the circle and the point up 1 and left 2 so that the center of the circle is now at the origin. This is just a translation of everything, so distances and slopes are preserved.

Drop a perpendicular from A to the x-axis. You can find the angle between the line connecting the circle center (origin) and A to the x-axis using inverse tangent.

Find the angle between the tangent line and the line connecting the circle center and A. You can do this by making a right triangle with the radius and using inverse sine.

Extend the tangent line so that it hits the x-axis. You should be able to use the information you have to find the angle between the tangent line and the x-axis.

5. May 23, 2008

### tiny-tim

Welcome to PF!

Hi JBD2! Welcome to PF!

Another way of finding the tangent lines through A is to write the equation for a general line through A, and then write the equation for where that line meets the circle.

It will meet the circle in only one point if it is a tangent!

That will be a quadratic equation, and you want the lines for which there are two identical roots.

6. May 23, 2008

### sennyk

This is going to get a little hairy, but here's the way I would approach the problem.

Your line goes through two points: $(11,8) (x,y)$

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
$$m = \frac{dy}{dx}$$

Once you set those two equal to one another, you have two equations (circle and slope) and two unknowns. You can solve that system using a CAS and have your answers for $(x_n,y_n)$. From there finding the equation of the line(s) is trivial.

7. May 23, 2008

### JBD2

Ok Thanks for the replies, I'll try that later, one more thing, this may sound stupid but what is a CAS?

8. May 23, 2008

### daveyinaz

Computer Algebra System.

9. May 23, 2008

### JBD2

What is dy and dx in m = dy/dx? Also what is the general line through A? Is that the middle line between the two tangents?

10. May 24, 2008

### tiny-tim

HI JBD2!
Have you done calculus yet?

If not, it's the gradient (the slope) of the line.

So, for ax + by +c = 0, or y = -(ax + c)/b it would be -a/b.
No, that would be the line with the equation (y - 8) = p(x - 11).

Obviously, for any p, this goes through (11,8), and its slope is p.

So as you change p, you rotate the line about (11,8), and get all the lines through (11,8).

11. May 24, 2008

### JBD2

I haven't done calculus yet, that's not until next year, and this question is supposed to be a bonus question so I'm a little confused

12. May 24, 2008

### tiny-tim

It's ok … you don't need calculus …

… just do it the way Tedjn or I suggested.

13. May 24, 2008

### JBD2

Using the way Tedjn suggested I figured out that the angle between the x-axis and the tangent was 64.77°. How do I get the slope of the line with this angle? Thanks for all the help.

EDIT: Wait, I just go tan(64.77) right? Making the slope 2.12?

Last edited: May 24, 2008
14. May 24, 2008

### JBD2

If my above assumption is right, then that would make the line y = 2.12x - 15.32, but how would I get the equation of the other tangent?

15. May 24, 2008

### tiny-tim

Hi JBD2!

I'm rather confused. You need to show some working if you want us to check what you've done.

Actually, now that I've re-read your original post, I see that you were almost there.

You correctly got all three sides of the triangle.

Now get the angle from those figures, and then both add and subtract it from the angle of the line to the centre (which obviously is 45º).

16. May 24, 2008

### JBD2

What I've done so far, is I moved the circle and point up 1 y value and to the left 2 x values so the circle is at 0,0 and the point instead of 11,8 is at 9,8. I drew a perpendicular line coming from point A (9,8) to the x-axis to make a right triangle between one of the tangents and the x-axis. I then did what Tedjn suggested and found the angle between the line connecting the circle center (origin) and A to the x-axis using inverse tangent. This turned out to be 48.36 degrees. Because of inverse tan(9/8). Then I found the angle between the tangent line and the line connecting the circle center and A by making a right triangle with the radius and using inverse sine. This angle turned out to be 23.13 degrees. I subtracted 23.13 from 48.36 to get 25.23 degrees in the right triangle with the tangent as a side and the perpendicular line as the other side. I subtracted angles of this triangle (180-90-25.23=64.77) to get 64.77 degrees between the x-axis and the tangent. I then did tan(64.77) to get 2.12 which would be the slope. To get the y-intercept or "b", I put (11,8) into the equation and solved for b so from:
y = 2.12x + b
8 = 2.12(11) + b
b = -15.32
So the final equation of that line was y = 2.12x - 15.32. If I did something incorrectly let me know. I just don't know how to do the final line. Thanks.

17. May 25, 2008

### JBD2

Does this look right?

18. May 25, 2008

### tiny-tim

Hi JBD2!
No, it's at 9,9.
Yes.
You should both add and subtract, to get both the tangents!

(and of course, it's not 48.36)
But you already had the angle between the x-axis and the tangent. This is the angle between the y-axis and the tangent.

19. May 26, 2008

### HallsofIvy

Staff Emeritus
I'm coming into this late but one method is that used by Fermat and DeCartes, "pre- Newton and Leibniz":

Any line through (11, 8) has the equation y- 8= m(x- 11) or y= mx- 11m+ 8. Put that into the equation of the circle, (x-2)² + (y+1)² = 25, and you get a quadratic equation for x, involving the parameter m. A quadratic equation may have 2, 1, or 0 solutions just as a line may cross a circle, twice, once, or not at all. The tangent line "crosses" the circle at only one point. Solve the quadratic equation using the quadratic formula. For what values of m does that have only one solution? You will find two such values for m since there will be two tangents.

20. May 26, 2008

### tiny-tim

Hi HallsofIvy!

Yes, I suggested that in post #5 … but you and I seem to be the only two people who like that idea!

(and I think the determinant comes out as a quartic in m unless the point is on one of the coordinate axes)