Slope of the tangents to the circle $x^2+y^2-2x+4y-20=0$

Solution 1:

Yes, your approach is correct. There is no need to convert to standard form though. From $$x^2+y^2-2x+4y-20=0$$ you can directly write $$2x + 2y\frac{dy}{dx} - 2 + 4\frac{dy}{dx} = 0\\ \implies \frac{dy}{dx} = \frac{1-x}{y+2}$$ for $y\ne -2$. Can you see what happens when $y = -2$?

Edit: For $y = -2$, let us consult the following graph of the circle. At $y = -2$, the circle has two vertical tangents (i.e. infinite slope). So, the slope $m = \infty$ at these points.

This should also be clear from the fact that $$\lim_{y\to -2} \frac{1}{|y+2|} = \infty$$

Solution 2:

Yes absolutely. After that $$\frac{dy}{dx}=-\frac{x-1}{y+2}.$$ Now at any point $(a,b)$ with $b\ne 2$ on the circle you get the gradient $$\frac{dy}{dx}_{x=a,y=b}=-\frac{a-1}{b+2}=m(say).$$ Now find the equation of the circle by $y=mx+c$.