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Problem: Find a tangent lines to the cirlce that tangent lines cross (4,1). $$x^2+y^2+4x-2y-11=0$$ I try to find $y'$ and I get $\frac{-x-2}{y-1}$. And idea after that was using $y-y_0=\frac{-x_0-2}{y_0-1}(x-x_0)$.

Ian Miller
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josf
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  • See also : http://math.stackexchange.com/questions/543496/how-to-find-the-equation-of-a-line-tangent-a-circle-and-a-given-point-outside-of – lab bhattacharjee Apr 01 '16 at 16:18

2 Answers2

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You working out is fine. Substitute in the values for $(x_0,y_0)$ and you should will notice you get an infinite gradient. Have a think about what that means.

It is a vertical line. So the answer is simply $x=4$.

Ian Miller
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\begin{align} x^2 + y^2 + 4x - 2y - 11 &= 0\\[0.3cm] 2x + 2yy' + 4 - 2y' &= 0\\[0.3cm] (2y-2)y' &= -2x-4\\[0.3cm] y' &= -\frac{x+2}{y-1} \end{align} Looks like your $y'$ is correct!

Next, plug $(x,y) = (4,1)$ into $y'$ to find the actual slope.. Ah, but we can't let $y = 1$ because then we get $0$ in the denominator. So, tell me, what does this mean about the tangent line at that point?

  • Why the downvote? –  Apr 01 '16 at 13:43
  • It might have been based on your first answer which didn't address the problem the OP was having with the evaluation of the gradient. – Ian Miller Apr 01 '16 at 13:44
  • The downvote came quite a bit after that, though, unless it just took a while for me to be notified. Meh, oh well. Haters gonna hate. –  Apr 01 '16 at 13:46