Find the equations of the normal and tangent. I'm having some trouble with this question:

Find the equations of the normal and tangent at the point $(-1, 3)$ on
  the circumference of a circle with a centre at $(4, 0)$.

I don't really know how to begin to attempt this question.
 A: A circle with center $(4,0)$ has equation $$(x-4)^2+y^2=r^2$$ To find $r$ plug in the point $(-1,3)$ and get $25+9=r^2=34$. The equation of the circle is $$(x-4)^2+y^2=34$$A line tangent to a circle is perpendicular to the radius. The radius passes through the points $(4,0)$ and $(-1,3)$. The slope of the radius' line segment is $$\frac{0-3}{4+1}=-\frac35$$ So the slope of the tangent line is $\dfrac53$ and it passes through the point $(4,0)$ so the equation of the tangent line is $$y=\frac53(x-4)$$. To find the normal line just find the equation of the line segment mentioned above.
A: Drawing a diagram will help you to see the statements that I make here. When stuck on a problem, always draw a diagram!
HINTS:
The normal line through a point on a circle also goes through the center of the circle. So just find the equation of the line that goes through the points $(-1,3)$ and $(4,0)$.
The tangent line through a point on a circle is perpendicular to the diameter through that point. So just find the equation of the line through $(-1,3)$ that is perpendicular to the normal line that you just found.
A: One approach is write an equation of any line going through $(−1,3)$ and the equation of this circle and, using discriminants ($ \Delta$ in a quadratic equation), calculate when those two equations have exactly one solution in common. 
Then, the line will be the tangent and the normal will be easy to calculate cause it's perpendicular to the tangent.
A: the equation of circle is
$$(x-4)^2+y^2=r^2$$
$$2(x-4)+2yy'=0$$
$$y'=\frac{4-x}{y}$$
at point (-1,3)
$$y'=\frac{4+1}{3}=\frac{5}{3}$$
the equation of tangent line is
$$y=\frac{5}{3}x+a$$
use the point (-1,3) to find $a$
the slope of normal is 
$$m=-\frac{3}{5}$$
so, the equation of normal line is
$$y=-\frac{3}{5}x+b$$
use the point (-1,3) to find $b$
