Equation for lines that are tangents to a circle

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Homework Statement


Find a differential equation whose solution is a family of straight lines that are tangents to the circle x^2+y^2=a^2 where a is a constant.

The Attempt at a Solution


So actually I'm stuck on the first part, coming up with such an equation. After some work I came up with
y=\pm(\frac{b(x+b)}{\sqrt{a^{2}-b^{2}}}+\sqrt{a^{2}-b^{2}})
(b varies from -1 to 1 to produce the different straight lines)
which reduces to
y=\pm\frac{bx+a^{2}}{\sqrt{a^{2}-b^{2}}}
and finding a differential equation whose solution is this family of straight lines is making my head hurt. Before I keep working I want to make sure this looks right. Graphing the equation works, but perhaps I'm misunderstanding the question.
 
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This is different to the equation I obtain. I first started by differentination of the circle equation to obtain:
<br /> \frac{dy}{dx}=-\frac{\alpha}{\beta}=\pm\frac{\alpha}{\sqrt{a^{2}-\alpha^{2}}} <br />
The above is the tangent of the line hitting the circle at the point (\alpha ,\beta ), now what you have to do is construct the line with the above tangent vale to get the equation of the tangent line.

They could just mean, differentiate the equation of a circle twice?
 
The answer is
y=xy&#039;\pm a\sqrt{(y&#039;)^2+1}
I wish I could just differentiate twice cause that makes things so much easier but I can't.

I used the equation you got as the slope for my line. That gives you a line that has the right slope but it doesn't intersect the circle at the right spot. The extra stuff in my equation moves the line around so it intersects the circle when their slopes are equal. I guess what I should be asking is, is that necessary? I'll try using the equation without extra stuff attached to it and see what I get.

Edit: Never mind, I got it, the formula was right, I was just going at it with the wrong approach. I kept squaring both sides to get rid of the ± because of some phobia I have of them, and once I stopped doing that, the problem became much simpler. I realized this after six pages of work...
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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