Answer
Center: $(1,0)$ $;$ Radius: $4$
Work Step by Step
$x^{2}-2x+y^{2}-15=0$
Take $15$ to the right side:
$x^{2}-2x+y^{2}=15$
Group the terms with $x$ together:
$(x^{2}-2x)+y^{2}=15$
Complete the square for the expression inside parentheses by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside and also on the right side of the equation.
In this case, $b=-2$
$\Big[x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}\Big]+y^{2}=15+\Big(-\dfrac{2}{2}\Big)^{2}$
$(x^{2}-2x+1)+y^{2}=15+1$
$(x^{2}-2x+1)+y^{2}=16$
Factor the expressions inside the parentheses, which is a perfect square trinomial:
$(x-1)^{2}+y^{2}=16$
This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$
It can be seen from the obtained equation that $(h,k)=(1,0)$, so the center of the circle is the point $(1,0)$
The radius is:
$r^{2}=16$
$\sqrt{r^{2}}=\sqrt{16}$
$r=4$
The graph is:
