Find a tangent lines to the circle that tangent lines cross (4,1). $x^2+y^2+4x-2y-11=0$

Question

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 $y-y_0=\frac{-x_0-2}{y_0-1}(x-x_0)$.

Answer 1

\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?

Answer 2

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 $x=4$.