Answer
$a)$ This equation represents a point
$b)$ There is no need to find the center or radius. There is also no need to sketch a graph
Work Step by Step
$x^{2}+y^{2}-6x-10y+34=0$
Take $34$ to the right:
$x^{2}+y^{2}-6x-10y=-34$
On the left side, group the terms with $x$ together and the terms with $y$ together:
$(x^{2}-6x)+(y^{2}-10y)=-34$
Complete the square for each group. To do so, add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and do it for both groups. For the first group, $b=-6$ and for the second group, $b=-10$
$\Big[x^{2}-6x+\Big(\dfrac{-6}{2}\Big)^{2}\Big]+\Big[y^{2}-10y+\Big(\dfrac{-10}{2}\Big)^{2}\Big]=-34+9+25$
$(x^{2}-6x+9)+(y^{2}-10y+25)=0$
Now, each group is a perfect square trinomial. Factor them both:
$(x-3)^{2}+(y-5)^{2}=0$
Since the part of the equation that represents the radius of the circle is $0$, then this equation represents a point.
$b)$
Since the original equation represents a point, there is no need to find the center or radius. There is also no need to sketch a graph