Slope of the tangents to the circle $x^2+y^2-2x+4y-20=0$ Find the slope of the tangents to the circle
$x^2+y^2-2x+4y-20=0$.
After I arranged into a standard form, which is $(x-1)^2 +(y+2)^2=25$
Centre point is $(1,-2)$ radius is $5$ unit.
Do I need to do the implicit differentiation?
If I did so,
The equation should be, $(2x-2)+ 2(y+2)\frac{dy}{dx}=0$.
Am I right?
 A: 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 $$\lim_{y\to -2} \frac{1}{|y+2|} = \infty$$
A: 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 $y=mx+c$.
A: Well I know you can find the slope of the tangent by differentiating the equation so will not be adding that but let me add that will make you feel the imagination of properties of the circle: The tangent to a circle is perpendicular to the radius of the circle
So, If $(x-h)^2 + (y-k)^2 = r^2$ is the equation of a circle and you want to find the slope of the tangent at point $P(a, b)$

*

*Find the  slope of the radius $\frac {b-k}{a-h}$
but if the tangent is perpendicular to the radius then the slope of the tangent will be $-\frac {a-h}{b-k}$
Or else simply use differentiation:
$2(x-h) + 2(y-k)y' = 0 \implies y' = -\left|\frac {x-h}{y-k}\right|_{(x, y)=(a, b)}$
