
#1
May2208, 11:49 PM

P: 62

Circle: (x2)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
May2308, 12:05 AM

P: 740

An easier way to do the problem may be to consider angles. What is the angle between the xaxis 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 xaxis? What is m = y/x in terms of that angle?




#3
May2308, 12:09 AM

P: 62

I'm not quite sure how I would find that angle though, because how would I know that the radius lines up with the xaxis? 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
May2308, 01:54 AM

P: 740

How can I find the equation of tangent lines from a circle to point A?
The radius doesn't have to line up with the xaxis. 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 xaxis. You can find the angle between the line connecting the circle center (origin) and A to the xaxis 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 xaxis. You should be able to use the information you have to find the angle between the tangent line and the xaxis. 



#5
May2308, 03:42 AM

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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
May2308, 08:23 AM

P: 73

This is going to get a little hairy, but here's the way I would approach the problem.
Your line goes through two points: [itex](11,8) (x,y)[/itex] [tex] m = \frac{y_2  y_1}{x_2  x_1} [/tex] [tex] m = \frac{dy}{dx} [/tex] 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 [itex](x_n,y_n)[/itex]. From there finding the equation of the line(s) is trivial. 



#7
May2308, 05:45 PM

P: 62

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




#8
May2308, 06:05 PM

P: 225

Computer Algebra System.




#9
May2308, 08:09 PM

P: 62

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
May2408, 10:21 AM

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HI JBD2!
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. 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
May2408, 12:18 PM

P: 62

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
May2408, 12:33 PM

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… just do it the way Tedjn or I suggested. 



#13
May2408, 05:11 PM

P: 62

Using the way Tedjn suggested I figured out that the angle between the xaxis 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? 



#14
May2408, 05:19 PM

P: 62

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
May2408, 06:27 PM

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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 reread 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
May2408, 09:33 PM

P: 62

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 xaxis to make a right triangle between one of the tangents and the xaxis. I then did what Tedjn suggested and found the angle between the line connecting the circle center (origin) and A to the xaxis 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 (1809025.23=64.77) to get 64.77 degrees between the xaxis and the tangent. I then did tan(64.77) to get 2.12 which would be the slope. To get the yintercept 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
May2508, 06:08 PM

P: 62

Does this look right?




#18
May2508, 06:33 PM

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Hi JBD2!
(and of course, it's not 48.36) 


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