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# Tangent to Circles? watch

1. More P3 (just a general question this time though)!!!

What methods can I use to prove that a given linear equation is a tangent to a circle (for which I have the equation)???

I thought about finding the gradient of the circles radius through implicit differentiation and ensuring that it was the normal to the tangent but due to the fact that I do not have the point of cantact, I cannot find the gradient of the radius.
2. (Original post by Leekey)
More P3 (just a general question this time though)!!!

What methods can I use to prove that a given linear equation is a tangent to a circle (for which I have the equation)???

Wouldnt you do summit like, the gradient of the tangent equation is equal to the dy/dx of the circle equation.

Because once you find dy/dx of circle equation, you put the x value into it and you get a gradient. This gradient is used in the tangent equation and if its the same as the gradient in the linear equation, then this proves that the linear equation is a tangent.
3. Let's assume y = mx is the tangent's equation, and x² + y² - 8x - 6y + 19 = 0 is the circle's equation.

Substitute y = mx is the circle's equation, then find b²-4ac. If it equals zero, that means the line is a tangent. This could also be used to find the value of m, simply equate b²-4ac to zero, and do some algebra.

4. (Original post by Leekey)
More P3 (just a general question this time though)!!!

What methods can I use to prove that a given linear equation is a tangent to a circle (for which I have the equation)???

I thought about finding the gradient of the circles radius through implicit differentiation and ensuring that it was the normal to the tangent but due to the fact that I do not have the point of cantact, I cannot find the gradient of the radius.
You have a line with equation
y = mx + c

And a circle, with equation
(x-a)^2 + (y-b)^2 = r^2

Sub y = mx+c into the circle equation and then you have a quadratic in x. Show that b^2 = 4ac, so there is only one root, and therefore only one point of contact - ie a tangent.
5. (Original post by shift3)
Let's assume y = mx is the tangent's equation, and x² + y² - 8x - 6y + 19 = 0 is the circle's equation.

Substitute y = mx is the circle's equation, then find b²-4ac. If it equals zero, that means the line is a tangent. This could also be used to find the value of m, simply equate b²-4ac to zero, and do some algebra.

This was the other method which I attempted but in the question that I encountered it simplified to 2x^2 so I thought that it was incorrect. Does it matter that "b" and "c" are effectively 0?
6. (Original post by BlueAngel)
you put the x value into it and you get a gradient.
Unless I can construct an infinite number of right-angles triangles in my minds eye, I cannot substitute a x value in when I do not know the point of intersection (i.e. the co-ordinates that would give a numerical valuse for the gradient of the radius) because I am not given it and I cannot just make it up!!!
7. (Original post by Leekey)
This was the other method which I attempted but in the question that I encountered it simplified to 2x^2 so I thought that it was incorrect. Does it matter that "b" and "c" are effectively 0?
I don't think a circle can have b=0... I'm not sure though.
8. (Original post by Leekey)
Unless I can construct an infinite number of right-angles triangles in my minds eye, I cannot substitute a x value in when I do not know the point of intersection (i.e. the co-ordinates that would give a numerical valuse for the gradient of the radius) because I am not given it and I cannot just make it up!!!

Sorry, God I feel so thick now. I mean I revised all week for P1 and P3 and I cant believe I got this wrong, urghhhh
Im not sure, but couldnt you use the centres coordinates of the circle?
9. (Original post by shift3)
I don't think a circle can have b=0... I'm not sure though.
This is why I was grossely confused!!! Im pretty sure my arithmetic was correct (can check it now because my friend has taken it to try at home). I was just really irritated that it was the only thing I couldn't do on P3 tonight!!!

So are we pretty sure that substituting the linear equation back into the forumla for the circle and the establishing the discriminant is the way to go with questions of this type?
10. (Original post by BlueAngel)
Sorry, God I feel so thick now. I mean I revised all week for P1 and P3 and I cant believe I got this wrong, urghhhh
Im not sure, but couldnt you use the centres coordinates of the circle?
No problem, Im having the same overload of maths problem!!!
11. (Original post by Leekey)
So are we pretty sure that substituting the linear equation back into the forumla for the circle and the establishing the discriminant is the way to go with questions of this type?
100%
12. (Original post by shift3)
100%
Cheer!!! Just godda hope that comes up on Wednesday now!!!
13. I hope we get anything as long as it's unlike June 2003's paper.
14. (Original post by shift3)
I hope we get anything as long as it's unlike June 2003's paper.
I did that for a mock and got 65/75 (raw mark) but I get some of the supposedly easy ones and get 48/75. I just want a 5 mark question on the binomial and a 70 mark question on integration by parts. If I get that then I will be a happy bunny!!!
15. (Original post by Leekey)
I did that for a mock and got 65/75 (raw mark) but I get some of the supposedly easy ones and get 48/75. I just want a 5 mark question on the binomial and a 70 mark question on integration by parts. If I get that then I will be a happy bunny!!!
Haha, a 70 mark integration by parts question?

That does sound like a good test combo as it doesnt have vectors.
16. (Original post by shift3)

That does sound like a good test combo as it doesnt have vectors.
I'll only let them include vectors if it goes something like this...

1a) Find equation of line through A and B
1b) Line AB and this line intersect....find point of intersection
1c) Find angle between the two lines and decide if they are perpedicular

17. Brilliant!
18. I haven't read all of the answers to this, but I'd form similtaneous equations, and when you solve them (by substituting the equation of the line into the equation of the circle), you should end up with a quadratic, with a repeated root.

The repeated root shows that the line is a tangent to the circle.
Two distinct roots would show that the line was a chord in the circle.
No real roots would show that they don't cross at all.

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