Tangent problem

  • #1

Homework Statement


Find the equation of the tangent drawn from the point(-5,4) to the circle x^2+y^2-2x-4y+1=0


Homework Equations


y-y1=m(x-x1)
x^2+y^2+2gx+2fy+c=0


The Attempt at a Solution


The equation of the tangent using the point is as follows y-4=m(x+5). Now, if I substitute the value of y in the given equation of the circle the equation gets messy as there will be two variables m and x in the quadratic equation. Is there any elegant way to solve the slope of the tangent line without having to go through the trouble of substituting the value of y into the given equation of the circle..
 

Answers and Replies

  • #2
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Homework Statement


Find the equation of the tangent drawn from the point(-5,4) to the circle x^2+y^2-2x-4y+1=0


Homework Equations


y-y1=m(x-x1)
x^2+y^2+2gx+2fy+c=0


The Attempt at a Solution


The equation of the tangent using the point is as follows y-4=m(x+5). Now, if I substitute the value of y in the given equation of the circle the equation gets messy as there will be two variables m and x in the quadratic equation. Is there any elegant way to solve the slope of the tangent line without having to go through the trouble of substituting the value of y into the given equation of the circle..

Have you drawn a sketch of the point and the circle? A sketch would eliminate half of your work.
 
  • #3
44
1
This is a typical question in analytic geometry. I believe the textbook must have covered it. There are at least 3 ways to solve it.
 
  • #4
Yes, I have drawn a sketch of the point and the circle on your suggestion. I managed to find out one slope of the tangent which is 0. there are two slopes according to the answer index of the book. One is 0 and the other is -0.75
 
  • #5
Mentallic
Homework Helper
3,798
94
If you substitute the tangent line into the circle equation (which you should have converted into the general form of a circle) then yes, you'll have a mess in x and m, but you can treat it as a quadratic in m with x being a constant, and then solve for x with the quadratic formula. But you don't even need to solve for x, all you need is to do something with the discriminant in the quadratic formula.
 
Last edited:
  • #6
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92
Isn't the question a bit confusing? There can be two tangents from the given point. Finding one of them is quite easy, it can be easily deduced from a sketch. @OP: Are there any options given with the problem?
 
  • #7
No, just the given point and the equation of a circle are given. I think there is a bit of a mistake in the question given in our book. They should have mentioned two tangents and both of them pass through the same point(-5,4) and are tangents to the circle having two different slopes given the fact that there are two different answers. Correct me if I am wrong
 
  • #8
Mentallic
Homework Helper
3,798
94
No, just the given point and the equation of a circle are given. I think there is a bit of a mistake in the question given in our book. They should have mentioned two tangents and both of them pass through the same point(-5,4) and are tangents to the circle having two different slopes given the fact that there are two different answers. Correct me if I am wrong

It depends on what's expected of your class and of the question. Is it enough for the question to expect to see if you can deduce that the max height of the circle is at y=4, hence the gradient of the line is 0? If not, it may expect you to find the more challenging gradient.
 
  • #9
ehild
Homework Helper
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How would you draw the tangent lines using Thales' circle? You can follow the same procedure. Find the centre and the radius of the Thales' Circle from the given coordinates of point P and the coordinates of the centre of the circle. Write up the equation of the Thales' circle and find the intersections A and B with the given circle. One intersection and P are two points of a tangent line.

ehild
 

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  • #10
maybe, I can find out the other tangent by using thales' circle. I also don't need to find it's slope. It is worth a shot after all. Thanks, ehild
 
  • #11
ehild
Homework Helper
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You can find it, it is easy:smile:

ehild
 

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