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The circle $x^2+y^2-4y+3=0$ passes through the points $(0,1),(-\frac{24}{25},\frac{43}{25}),(1,2)$. Its Centre is $(0,2)$ and its radius is $1$. I am asked to find the tangent to the circle through the point $(1,2$). This seems like a trivial problem but I have hit something which is not clear to me. I write the equation of the tangent as $y-2=m(x-1)$ which gives $mx-y+2-m=0$ as the tangent. I then calculate the perpindicular distance from the centre $(0,2)$ to the tangent and equate this distance to the radius $1$. This gives $$\frac{m(0)-1(2)+(2-m)}{\sqrt{(1+m^2)}}=1$$ which in turn gives $$-\frac{m}{\sqrt{(1+m^2)}}=1$$ which gives $$m^2=1+m^2$$

I suppose its correct if $m = \infty$ (which it is as the tangent is $x=1$) but this answer does not roll out of the equations. What have i done wrong here?

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  • $\begingroup$ Divide $m^2=1+m^2$ by $m^2$ to get $1/m=0$, thereby from the equation of the tangent, $(y-2).1/m=x-1$, $x=1$ is the required tangent. Does it answer your question..? $\endgroup$ – Tapu Mar 11 '13 at 11:54
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It looks like you've already figured out what happened: You set up the equation for the tangent in a way that excluded a vertical tangent; then because the tangent is in fact vertical you got an equation that isn't satisfied by any finite slope; then you noticed that you can let it make sense nevertheless by allowing for an infinite slope. I'm not sure what you mean by "this answer does not roll out of the equations". Either you assume finite slopes; then what you did wrong was to exclude vertical lines, and the tangent happened to be vertical; or you allow infinite slopes; then the answer does roll out of the equations. I don't think there's much more to say about it than that.

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  • $\begingroup$ Yes this answers my question, many thanks, also your explaination of how i prevented myself from solving the problem by excluding a vertical tangent is clear. What I meant by"rolling out of the equation" was that I expected to get a value(s) of m to plug back into the line equation to give me the tangent. I failed to see the solution for m^2=1+m^2 which you showed above. $\endgroup$ – twa14 Mar 11 '13 at 13:23

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