equation of tangent to a circle with zero gradient Find the equation of the tangent to the circle $$(x-3)^2 + (y-2)^2 = 1$$ at the point $(4,2)$.
I cannot show the working of the tangent equation as gradient/slope = 0.
So how can I find the gradient of tangent?
I think it is also obvious that the tangent is $x = 2$ but what about working?
The general form for a line is $ax+by = c$. This form more generally describes lines than the form $y=mx+b$, simply the functional implication of the latter form requires us to use a point at infinity in the parameterization of vertical lines.

So, a vertical line at $x=1$ can be described by $1\cdot x+0\cdot y = 1$, represented by a point in parameter space as $(a,b,c)=(1,0,1)$.

The tangent to your circle at $(x,y) = (4,2)$ has $x=4$ everywhere, so can be written as a point in parameter space as $(1,0,4)$.

For more detail, see here: http://www.mathsisfun.com/algebra/line-equation-general-form.html
Since when the line x=2 started having zero gradient, that too obviously!
Answer is y=1 and y=3.
It can be seen very easily from graph.
Mathematically any line with 0 slope is y=constant. So put y=k in equation and use condition of tangency.